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Class Q/4 531 
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COMPLETE TRIGONOMETRY 



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i 



BY 



WEBSTER WELLS, S.B. 

PROFESSOR OF M A.THEMATICS IN THE MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY 



3>0:^C 



BOSTON, U.S.A. 

D. C. HEATH & CO., PUBLISHERS 

1900 



40355 

IJt>r«(ry of Congress 

Two CofiES Received 
AUG 30 1900 

AUG 30 19U0 
FIRST copy. 

2114 Co^l Odiverodi t* 

ORDER DlVtSlOH 

SFP n 1900 



Q A r 3 1 



Copyright, 1900, 
By WEBSTER WELLS. 



r->\v PREFACE. 



The present volume is a revision of the author's Essen- 
tials of Trigonometry. 

In preparing the new edition, many improvements have 
been effected; the attention of teachers is specially invited 
to the following features of the work : 

1. The proofs of the functions of 120°, 135°, 150°, etc.; 
§ 27. 

2. The proofs of the functions of (-A), and (90° +^), 
in terms of those of ^; §§28, 29. 

3. The method of solution employed in the examples of 
§§ 33 and 34. 

4. The general demonstration of the formulae 

tan X = -, and sin^a; + cos^o? = 1, 

COSOJ 

in §§ 36 and 38; the four cases being considered together. 

5. The general demonstrations of the formulae 

cot a; = — — , sec^o^ = 1 + tannic, and csc^x = 1 + cot^a; 
since 

in §§ 37 and 39. 

6. The proofs of the formulae for sin {x-{-y) and cos (x-{-i/), 
when X and y are acute ; the two cases when x-\- y is acute 
or obtuse being considered together ; § 41. 

7. The proofs of the formulae for tan i x and cot i aj ; § 48. 

8. The illustrative examples in § 49. 



•IV 



Preface. 



9. The solution of right triangles by Natural Functions ; 
see Ex. 1, page 66. 

The new work contains a great many more examples than 
the old ; they have been selected with great care, and most 
of them are new. It is not expected that every class will 
solve all the examples ; they are sufficiently numerous to 
furnish a variety in successive years. 

Attention is specially invited to the sets in §§ 96, 114, 157, 
and 160. 

In § 112 will be found a set of miscellaneous examples in 
the solution of plane oblique triangles, and in § 155 a set in 
spherical oblique. 

The results have been worked out by aid of the author's 
New Four Place Logarithmic Tables, which contain also 
Tables of Natural Functions. 

WEBSTER WELLS. 

Massachusetts Institute of 
Technology, 1900. 



CONTENTS. 

PLANE TRIGONOMETRY. 

PAGE 

I. Trigonometric Functions of Acute Angles . 1 

11. Trigonometric Functions of Angles in General 7 

III. General Formula 26 

IV. Miscellaneous Theorems 38 

V. Logarithms 50 

Properties of Logarithms 52 

Applications . . 58 

VI. Solution of Right Triangles . . . .65 

Formulae for the Area of a Right Triangle . . 72 

VII. General Properties of Triangles ... 75 

Formulae for the Area of an Oblique Triangle . 80 

VIII. Solution of Oblique Triangles .... 82 

SPHERICAL TRIGONOMETRY. 

IX. Geometrical Principles 97 

X. Right Spherical Triangles 101 

Solution of Right Spherical Triangles . . . 108 



vi Contents. 



PAGE 



XL Oblique Spherical Triangles .... 118 

General Properties of Spherical Triangles . . 118 

Napier's Analogies . . . . ^ . 124 

Solution of Oblique Spherical Triangles . . 127 

XII. Applications . . 139 

Formula : 

Plane Trigonometry . . 144 

Spherical Trigonometry . . . . . . . 147 

Answers 

Use of the Tables 



PLAICE TRIGONOMETRY. 



35*JC 



I. TRIGONOMETRIC FUNCTIONS OP ACUTE 
ANGLES. 

1. Trigonometry treats of the properties and measure- 
ment of angles and triangles. 

In Plane Trigonometry, we consider plane figures only. 

2. Definitions of the Trigonometric Functions of Acute 
Angles. 

B 




Let BAC be any acute angle. 

From any point in either side, as B, draw line BC per- 
pendicular to AC, forming right triangle ABC. 

We then have the following definitions, applicable to 
either of the acute angles A ov B: 

In any right triangle, 

The sine of either acute angle is the ratio of the opposite 
side to the hypotenuse. 

The cosine is the ratio of the adjacent side to the hypotenuse. 

The tangent is the ratio of the opposite side to the adjacent 



The cotangent is the ratio of the adjacent side to the oppo- 
site side. 



Plane Trigonometry. 

The secant is the ratio of the hypotenuse to the adjacent 
The cosecant is the ratio of the hypotenuse to the opposite 



We also have the following definitions : 

The versed sine of an angle is 1 minus the cosine of the 
angle. 

The coversed sine is 1 minus the sine. 

The eight ratios defined above are called the Trigono- 
metric Functions of the angle. 

Representing sides EC, CA, and AB, by a, b, and c, 
respectively, and employing the usual abbreviations, we 
have: 



• A ^ 

sm A = ~' 
c 


tan^ = -- 




b 


vers A = l 


b 
c 


A ^ 

i[ COS A = -' 
c 


C0t-4 = — 

a 


A C 

CSC ^ = — 

a 


covers A = l 


a 

c 


sinB = l. 


tan.B = -. 
a 




sec -B = — 

a 


vers B = l 


a 

c 


G0SB = -- 

c 


cot 5 = ?. 




b 


covers B = l 


b 
c 



3. It is important to observe that the values of the trigo- 
nometric functions depend solely on the magnitude of the 
angle, and are entirely independent of the lengths of the 
sides of the right triangle which contains it. 




For let B and B' be any two points in side AD of angle 
DAE, and draw lines BC and B'C perpendicular to AE. 



Trigonometric Functions. 



Then, by the definitions of § 2, 

sinA = , and sm^^ - — -• 

But right triangles ABC and AB'C are similar, since 
they have angle A common. 
Whence, by Geometry, 

BC^B'C 
AB AB'' 

Thus the two values found for sin A are equal. 
The same may be proved true of each of the remaining 
functions. 

4. We have from § 2, 

sin A = cos B. sec A = esc B. 

tan A = cot B. vers A — covers B. 

But B is the complement of ^. 

Hence, the sine, tangent, secant, and versed sine of any 
acute angle are, respectively, the cosine, cotangent, cosecant, 
and coversed sine of the complement of the angle. 

5. From § 2, sin ^ = - = cos B, and cos J. = - = sin B. 

C G 

Whence, 

a = c sin A = c cos B, and h — c sin B = c cos A. 

That is, in any right triangle, either side about the right 
{ingle is equal to the sine of the opposite angle, or the cosine 
of the adjacent angle, multiplied by the hypotenuse, 

6. To find the Values of the Other Seven Functions of an 
Acute Angle, when the Value of Any One is Given, 

1. Given csc^ = 3; find the values of the remaining 
functions of A. 



Plane Trigonometry. 
B 




We may write the equation esc A= 



Since the cosecant is the hypotenuse divided by the opposite side, 
we may regard A as one of the acute angles of right triangle ABC-, 
in which hypotenuse AB = 3, and opposite side BC =1. 



By Geometry, AC = ^ AB^ -BC'^ = \/9 - 1 = V8 = 2 V2. 



Then by the definitions of § 2, 
1 



sin J. = 



cos^ = 



3 

2\/2 



tan^ 



1 

2\/2 



cot ^ r= 2 \/2. 

1 2 



sec A = :2 

2V2 

vers A= 1 



2V2 
3 



2. Given vers A 



covers A 

2 



1 - 



3 3 
; find the value of cot A. 
B 




Since vers ^ = 1 — cos ^, we have cos A 



5 5 



Then, in right triangle ABC, we take adjacent side AC = S, and 
hypotenuse AB = 5. 



Whence, BC = VaB^ - ^C^ = V25 - 9 = VI6 = 4. 

3 

Then, by definition, cot A = -' 



Trigonometric Functions. 



EXAMPLES. 

In each of the following, find the values of the remaining 
functions : 



3. 


sin^ = ^. 


6. 


CSC A = 7. 


9. 


sec A=: X. 


4. 


vers ^ = A. 


7. 


cosA-'f. 
14 


10. 


tan A = -' 

X 


5. 


cot^ = ^. 


8. 


2 

covers A = -^« 


11. 


sm A = -' 





12. Given cot ^ = - j find sin A. 



13. Given esc A 



41 

40 



find cos A. 



14. Given sec J. = 5 ; find cot A. 

21 

15. Given cos A = — ; find esc A. 

29' 

16. Given tan A = — ^ : find sec A. 



17. Given sinJ. = -: find vers A 



7. Functions of 45°. 




Let ABC be an isosceles right triangle, C being the right 
angle, and sides AC and BC being each equal to 1. 



Then, ZA=4.5°, and AB=-^AC'-i-BC'=-\/l+l = V2. 



6 Plane Trigonometry. 

Whence, by definition, 

sec45°=V2. 



sin45° = A: = +V2. 

V2 

cos45°- — = iV2. 

V2 ' 



csc45°=V2. 



vers 45° = 1-1 V2 



covers45° = l-iV2 




2-V2^ 
2 

2-V2 



Let ^57) be an equilateral triangle having each side equal 
to 2, and draw line AC perpendicular to BD. 

By Geometry, BC=\BD = 1, Z BAC =i Z BAD = 30°. 



Also, AC = ^AB'-BC' = V4.-1=^^. 

Then from right triangle ABC, by definition, 



sin 30° = ^ = cos 60°. 



cos 30°=^= sin 60°. 



tan 30^ = — = 1 V3 = cot 60°. cot 30° = V3 = tan 60= 

V3 ' 



sec 30^^ 



V3 



= 2 V3 = CSC 60°. CSC 30° = 2 = sec 60°. 



V3 



vers 30° = 1 - -^ = covers 60°. 

2 

covers 30° = 1 - - = - = vers 60°. 

2 2 



Trigonometric Functions. 



II. TRIGONOMETRIC FUNCTIONS OF 
. ANGLES IN GENERAL. 

9. In Geometry, we are, as a rule, concerned with angles 
less than two right angles ; but in Trigonometry it is con- 
venient to consider them as unrestricted in magnitude. 





Let be the centre, and XX' and YY' a pair of perpen- 
dicular diameters, of circle AY^ \ OY being above, and Y' 
below, XX', when OX is horizontal and extends to the 
right, and OX' to the left, of 0. 

Let radius OA (Fig. 1) start from the position OX, and 
revolve about point as a pivot towards the position OY. 

When OA coincides with OY, it has generated an angle 
of 90°; when it coincides with OX', of 180°; with OY', 
of 270°; with OX, its first position, of 360°; with OF again, 
of 450° ; and so on. 

Hence, a meaning may be attached to a positive angle of 
any number of degrees. 

10. We may also conceive of a negative angle of any 
number of degrees. 

Thus, if a positive angle indicates revolution from the 
position OX towards OY", a negative angle may be taken as 
indicating revolution from the position OX in the opposite 
direction, towards OY'. 



8 Plane Trigonometry. 

Thus, if radius OA' (Fig. 2) starts from the position OX, 
and revolves about point as a pivot towards the position 
Y', when it coincides with T\ it has generated an angle 
of - 90°; when it coincides with 0X\ of - 180°; with OT, 
of -270°; and so on. 

Note. It is immaterial which direction we consider the positive 
direction of rotation ; but having at the outset adopted a certain direc- 
tion as positive, our subsequent operations must be in accordance. 

11. In generating a positive or negative angle of any 
number of degrees, the line from which the rotation is sup- 
posed to commence is called the initial line of the angle, and 
the final position of the rotating radius the tei^minal line. 

12. To designate an angle, we always write first the letter 
at the extremity of the initial line. 

Thus, in designating the angle formed by the lines OX 
and OA, if we regard OX as the initial line, we should call 
it XOA ; and if we regard OA as the initial line, we should 
call it AOX. 

13. There are always two angles less than 360° in abso- 
lute value, one positive and the other negative, having the 
same initial and terminal line. 

Thus there are formed by OX and OA (Fig. 2) the posi- 
tive angle XOA between 270° and 360°, and the negative 
angle XOA between 0° and - 90°. 

We shall distinguish between these angles by referring to 
them as 'Hhe positive angle XOA,^^ and ''the negative 
angle XOJ.','' respectively. 

14. Rectangular Co-ordinates. 

Let XX^ and YY^ be two straight lines intersecting at 
right angles at O ; the letters being arranged as in the fig- 
ures of § 9. 

Let Pi be any point in the plane of XX' and YY\ and 
draw line P^M perpendicular to XX'. 



Trigonometric Functions. 9 

Then OM and P^M are called the rectangular co-ordinates 
of Pi ; OM is called the abscissa, and PiM the ordinate. 



a 



?i (b,.cO 



P3 (-&,-«) 



P4 (Py-Ci) 



The lines of reference, XX' and FF', are called the axis 
of X and the axis of Y, respectively, and is called the 
origin. 

It is customary to express the fact that the abscissa of a 
point is h, and its ordinate a, by saying that for the point 
in question x = h and y = a; or, more concisely, we may 
refer to the point as " the point (b, a)," where the first term 
in the parenthesis is understood to be the abscissa, and the 
second term the ordinate. 

15. If, in the figure of § 14, M and JSf be points on OX 
and OX', respectively, such that 0M= ON=b, and lines 
P1P4 and P2P3 be drawn through M and N, respectively, 
perpendicular to XX', making P^M = P^JSF = P.^JSF = P^M = a, 
each of the points Pi, P^, P3, and P4 will have its abscissa 
equal to b, and its ordinate equal to a. 

To avoid this ambiguity, abscissas measured to the right 
of are considered positive, and to the left, negative; and 
ordinates measured above XX' are considered positive, and 
below, negative. 

Then the co-ordinates of the points will be as follows : 



Pi, (b, a)', P„ (- b, a); P„ (- b, - a); P^ (b, - a). 



lo Plane Trigonometry. 

Note 1. It is understood, in the above convention with regard to 
signs, that the figure is so placed that OX is horizontal, and extends 
to the right from 0. 

Note 2. In all the figures of the present chapter, the small letters 
are understood as denoting the lengths of the lines to luhich they are 
attached, without regard to their algebraic signs; hence they always 
represent positive quantities. 

16. If a point lies upon XX', its ordinate is zero; and 
if it lies upon YY', its abscissa is zero. 

17. General Definitions of the Functions. 

We will now give general definitions of the trigonometric 
functions, applicable to any angle whatever. 

Construct axes in such a way that the initial line of the 
angle shall be the positive direction of the axis of X, and 
the vertex the origin. 

From any point in the terminal line drop a perpendicular 
to the axis of X, and find the co-ordinates of this point. 

Then, the sine of the arigle is the ratio of the ordinate of the 
point to its distance from the origin. 

The cosine is the ratio of the abscissa to the distance. 
The tangent is the ratio of the ordinate to the abscissa. 
The cotangent is the ratio of the abscissa to the ordinate. 
The secant is the ratio of the distance to the abscissa. 
The cosecant is the ratio of the distance to the ordinate. 

Note. The above definitions include those of § 2. 
The definitions of the versed sine and coversed sine, given in § 2, 
are sufficiently general to apply to any angle whatever. 

18. We will now apply the definitions of § 17 in the fol- 
lowing figures. 

In each case, we construct axes in such a way that the 
initial line of the angle shall be the positive direction of the 
axis of X, and the vertex the origin. 



Trigonometric Functions. 
I. Let XOP2 be any angle between 90° and 180°. 

Y 



I I 




Let P2 be any point on the terminal line, and draw P2M 
perpendicular to XX' ; let P2M = a, OM = b, and OP2 = c. 
Then the co-ordinates of P2 are {— b, a). 
Whence, by definition. 



sin XOPo 



tan XOP, 



sec XOPo = 



-b 



cos XOP, 



cot XOP. 



CSC XOP.2 



c 
-b 



h^_b_ 
c 



11. Let XOP3 be any angle between 180° and 270°. 




P3(-Z)-a) 



Let Pg be any point on the terminal line, and draw P^M 
perpendicular to XX' ; letP3lf=a, OM=b, and OP^ = e. 
Then the co-ordinates of P3 are (—b, — a). 



12 Plane Trigonometry 

Whence, by definition, 



tan XOP^ 



— a _a 



sec XOPo = 



cos XOP3 = 
cot XOP3 = 
CSC XOPo z= 



-b^ b 

c c - 

-b^b 

— a a 

_c__ _ c 

— a a 



III. Let XOP4 be any angle between 270° and 360°. 
Y 




PA'brO') 



Let P4 be any point in the terminal line, and draw P^M 
perpendicular to XX' ; let P^M = a, OM = b, and OP4 = c. 
Then the co-ordinates of P4 are (6, — a). 
Whence, by definition, 



sinX0P4 = — ^ = 

c 

tanX0P4-^^^ = 
b 

sec XOP4 = 



cos XOP4 



cotXOP4 



— a 



— a a 



19. It is evident that the terminal lines of any two 
angles which differ by a multiple of 360° are coincident, 
and hence the trigonometric functions of two such angles 
are identical. 



Trigonometric Functions. 



13 



Thus, the functions of 50°, 410°, 770°, - 310°, etc., are 
identical. 

20. If the initial line of an angle coincides with OX, and 
its terminal line lies between OX and Y, the angle is said 
to be in the Jlrst quadrant; if the terminal line lies between 
T and OX', the angle is said to be in the second quadrant ; 
if between OX' and OY', in the third quadrant; if between 
OF' and OX, in the fourth quadrant. 

Thus, any positive angle between 0° and 90°, or 360° and 
450°, or any negative angle between — 270° and — 360°, is 
in the first quadrant -, any positive angle between 90° and 
180°, or 450° and 540°, or any negative angle between — 180° 
and — 270°, is in the second quadrant. 

21. It follows from the definitions of § 17 that, for any 
angle in the first quadrant, all the functions are positive. 

It is also evident by inspection of the results of § 18 that : 

In the second quadrant, the sine and cosecant are positive, 
and the cosine, tangent, cotangent, and secant are negative. 

In the third quadrant, the tangent and cotangent are positive, 
and the sine, cosine, secant, and cosecant are negative. 

In the fourth quadrant, the cosine and secant are positive, 
and the sine, tangent, cotangent, and cosecant are negative. 

It is customary to express the above in tabular form, as 
follows : 



Functions. 


First 
Quad. 


Second 
Quad. 


Third 
Quad. 


Fourth 
Quad. 


Sine and cosecant . . . 
Cosine and secant . . . 
Tangent and cotangent . 


+ 
-f- 

+ 


+ 


+ 


+ 





14 Plane Trigonometry. 

22. Functions of 0° and 360°. 



a P(a.O\ 







The terminal line of 0° coincides with the initial line OX. 
Let P be a point on OX such that OP = a. 
Then by § 16, the co-ordinates of P are (a, 0). 
Whence by definition, 



sinO' 







0. 



tan 0' 







0. 



sec 0= 



1. 



cos 0° = - = 1. 



cotO° = ^-oo. 



cscO°=:- = oo. 



By § 19, the functions of 360° are the same as those of 0' 
23. Functions of 90°. 



T 






P(0,a) 




a 


f^90o 












For the angle 90°, OF is the terminal line. 
Let P be a point on OY, such that OP = a. 
Then the co-ordinates of P are (0, a). 
Whence by definition. 



sin 90° = - = 1. 
a 



cos 90^ 



0. 



tan 90° = - = 00. 




cot 90° = - = 0. 

a 



sec 90° = ^ = 00. 



CSC 90° = - = 1. 
a 



Trigonometric Functions. 
24. Functions of 180°. 



15 



.^^ 



180= 



P(-a,0) 



For the angle 180°, OX' is the terminal line. 
Let P be a point on OX', such that OP = a. 
Then the co-ordinates of P are (—a, 0). 
Whence by definition, 



sin 180° = - =0. 
a 



cos 180° = — ^ = - 1. 
a 



tan 180° = -^ = 0. 
— a 



cot 180° = ---= 00. 



sec 180° = -^ = - 1. 
— a 



CSC 180° = ^ =00. 



25. Functions of 270°. 



270° 



X- 




T (0,-a) 



For the angle 270°, OT' is the terminal line. 
Let P be a point on OT', such that OP = a. 
Then the co-ordinates of P are (0, — a). 



i6 



Plane Trigonometry. 



Whence by definition, 

sin 270° = ^^ = -!. 
a 

tan270°=-=p=:oo. 



cos 270° = - =0. 
a 



cot 270° = 



— a 



0. 



sec 270^ 







00. 



esc 270°=-^ = -!. 
— a 



Note. No absolute meaning can be attached to such a result as 
cot 0° = 00 ; it merely signifies that as an angle approaches 0°, its co- 
tangent increases without limit. 

A similar interpretation must be given to the equations esc 0° = cxj, 
tan 90° = GO, etc. 

26. Given the value of one function of an angles to find the 
values of the remaining functions. (Compare § 6.) 

3 

1. Given sin A — — -\ required the values of the remain- 
ing functions of A. 

The example may be solved by a method similar to that of § 6 ; 
since the sine is the ratio of the ordinate to the distance, v^e may 
regard the point of reference as having its ordinate equal to — 3, and 
its distance equal to 5. 

There are tvm points, P and P', which are 3 units below the axis 
of X, and distant 5 units from O. 




P(-4,-3) 



P'(4,-3) 



There are then two angles^ XOP and XOP', in the third and fourth 
quadrants, respectively, either of which may be the angle A. 



Now, OM = OM' = ^OP^ - PM^ = V25 -9 = 4. 

Then co-ordinates of Pare (—4, — 3), and of P', (4, — 3). 



Trigonometric Functions. 

Whence by definition : 



17 



Angle. 


Cos. 


Tan 


Co.. 


Sec. 


Csc. 


XOP 


4 
5 


3 
4 


4 
3 


5 
4 


5 
3 


XOP' 


4 

5 


_3 
4 


4 

3 


5 
4 


5 
3 



Thus the two solutions to the problem are : 

cosA=T -, tan A = ±-, cot A = ± -, sec ^ = =F -, csc ^ = — - ; 
5' 4' 3' 4' 3 

where the upper signs refer to XOP, and the lower signs to XOP'. 

2. Given cot A = 3-, required the values of tlie remain- 
ing functions of A. 

The equation may be written in the forms 

3 Q 

cot A = -, or cot A = — ^• 

We may then regard the point of reference as having its abscissa 
equal to 3 and its ordinate equal to 1, or as having its abscissa equal 
to — 3 and its ordinate equal to — 1. 

There are two angles, XOP and XOP', in the first and third quad- 
rants, respectively, either of which satisfies the given condition. 




P^(-3,-l) 



Then, OP = OP' = '^ 03f + P3f = \/9 + 1 = VlO. 



1 8 Plane Trigonometry. 

Whence by definition : 



Angle. 


Sin. 


Cos. 


Tan. 


Sec. 


Csc. 


XOP 


1 


3 


1 


VlO 


VlO 




VlO 


VlO 


3 


3 




XOP' 


1 


3 


1 


VTo 


-VlO 




VlO 


VlO 


3 


3 





Thus the two solutions are 



sin ^ = ± — ^, cos ^ = ± — ^, tan A = -., 
VlO VlO 3 



sec A = ± 



10 



3 



CSC ^ = ± VlO. 



Note. It must be clearly borne in mind, in examples like the 
above, that the "distance" is always positive. 

EXAMPLES. 

In each of the following, find the values of the remaining 
functions : 



3. secJ. = -- 
4 



7. csc ^ = — 



25 



4. cot^--— 8. tSinA = ~ 
5 40 



5. sin A 



15 



9. sec^ = 



21 



6. GOsA = — -—- 10. sin^ = — -• 
29 5 

27. Functions of 120°, 135°, 150°, etc. 



11. tanJ^=-7. 

12. csc ^ = 3. 



13. cos^ = -. 

b 

14. cot A = x. 




Trigonometric Functions. 



19 



Let 0PM be a right triangle having OP, OM, and PM 
equal to 2, 1, and VS, respectively, and Z POM =60°. 
(Compare § 8.) 

Then Z XOP = 120°, and co-ordinates of P are (-1, V3). 



Whence by definition, 




sin 120° = ^^• 


cos 120° = -^. 


tanl20° = -V3. 


cot 120°= ^ = ^V3. 
V3 3 


sec 120° = -2. 


CSC 120°= ^ =^V3. 
V3 3 



In like manner may be proved the remaining values given 
in the following table, which are left as exercises for the 
student : 



Angle. 


Sin. 


Cos. 


Tan. 


CoiJ. 


<Sec. 


(7w. 


120° 


iV3 


-4 


-V3 


-iV3 


-2 


fV3 


135° 


iV2 


-iV2 


-1 


- 1 


-V2 


V2 


150° 


1 


-iV3 


-iV3 


-V3 


-fV3 


2 


210° 


-i 


-iV3 


iV3 


V3 


-fV3 


-2 


225° 


-1V2 


-4V2 


1 


1 


-V2 


-V2 


240° 


-iV3 


-4 


V3 


|V3 


-2 


-fV3 


300° 


-iV3 


i 


-V3 


-iV3 


2 


-|V3 


315° 


-iV2 


1V2 


-1 


-1 


V2 


-V2 


330° 


-4 


iV3 


-1V3 


-V3 


|V3 


-2 



28. Functions of (—-4) in terms of those of A. 

To prove the formulm 



sin ( — J.) = — sin A, 
tan (— A^= — tan A, 
sec(— ^)= sec^, 
for any value of A. 



cos(— ^)= cos^, 
cot(— ^)= — cot^, 
CSC ( — J.) = — CSC A, 



(1) 



20 



Plane Trigonometry. 



There may be four cases : A in the first quadrant (Fig. 1), 
A in the second quadrant (Fig. 2). A in the third quadrant 
(Fig. 3), or A in the fourth quadrant (Fig. 4). 




Fig. 1. 




Tig. 3. 




In each figure, let the positive angle XOP represent the 
angle A, and the negative angle XOP' the angle — A. 

Draw PM perpendicular to XX', and produce it to meet 
OP' at P'. 

In right triangles 0PM and OP'M, side OM is common, 
2.rLd.Z.P0M=AP'0M. 

Then the triangles are equal, and PM=P'M and 
0P = 0P'. 

Therefore, in each figure, 

abscissa P' = abscissa P, 
ordinate P' = — ordinate P, 



and 



distance P' = distance P. 



Then, sin(— ^) = 
cos(— ^) = 
tan(— ^) = 
cot(- J) = 



Trigonometric Functions, 
ord. P' ord. P 



21 



dist.P'' 

abs. F 
dist.P' 

ord. P 
abs. P' 

abs.P 
ord. P' 



dist. P 

abs.P 
dist.P 

ord.P 
abs.P' 

abs. P 
ord.P' 



= — sin A, 



= cos A. 



— tan A. 



cot A 



/ .X dist.P dist.P > 

sec(— ^)=^ — ^= -^ — -= sec^. 



csc(— ^) = 



abs. P 



dist. P' 
ord. P' 



abs.P 



dist.P 
ord. P 



= — esc A 



29. Functions of (90° -f ^) in terms of those of A, 

To prove the formulce 



sin (90° + ^)= cos^ 
tan(90° + ^)=-cot^, 
sec(90° + ^) = -csc^, 

for any value of A. 
PC 



cos(90° + ^) = -sin^ 
cot(90° + ^)=-tan^, 
esc (90° + ^)= sec^, J 




(2) 




22 



Plane Trigonometry. 





I 



Fig. 3. 

There may be four cases : A in the first quadrant (Fig. 1), 
A in the second quadrant (Fig. 2), A in the third quadrant 
(Fig. 3), or A in the fourth quadrant (Fig. 4). 

In each figure, let the positive angle XOP represent the 
angle A, and the positive angle XOP' the angle 90° -f A. 

Take OP' = OP, and draw PM and P'M' perpendicular 
to XX'. 

Since OP is perpendicular to OP, and OJf to P'M\ 

ZPOM=ZOPM'. 

Then right triangles 0PM and OP'M' have the hypote- 
nuse and an acute angle of one equal, respectively, to the 
hypotenuse and an acute angle of the other, and are equal. 

Whence, PM= OM' and 0M= P'M'. 

Therefore, in each, figure, 

ordinate P = abscissa P, 

abscissa P = — ordinate P, 

and distance P' = distance P. 

ord. P' abs. P 



Then, sin (90° -\-A) = 
cos (90° -\-A) = 



dist. P 

abs. P' 
dist. P' 



dist. P 

ord. P 
dist. P 



= cos A. 
= — sin A. 



Trigonometric Functions. 23 



tan (90° + ^) = 2£d^' = _ 5]^ = - cot A 
~ ord. P 

cot (90° + A)= "":• " , = - ^^^^ = - tan A. 
V ^ y „„;] D, abs. P 



abs. 


P 


abs. 


P' 


ord. 


P' 


dist. P' 


abs. 


P' 


dist. P' 



sec(90° + ^) = T^'\ = -^^^r^ = -cscA 
V ^ ^ .u. Df ord. P 

csc(90° + ^)- ^T\ = distP^ gg^^^ 
^ ^ ord. P abs. P 

30. The results of § 29 may be stated as follows : 

The sine, cosine, tangent, cotangent, secant, and cosecant of 
any angle are equal, respectively, to the cosine, minus the sine, 
minus the cotangent, minus the tangent, minus the cosecant, 
and the secant, of an angle 90° less. 

31. Functions of (90° — A) in terms of those of A. 
By §30, sin (90°-^)= cos (-^) = cos ^ (§ 28). 

cos (90° - ^) = - sin (- ^) = sin A. 
tan (90° - ^) = - cot (- ^) = cot A. 
'" . cot(90°-^)=-tan(-^) = tanA 

sec (90° - ^) = - CSC (- ^) = CSC A. 
CSC (90° -A)= seG{-A) = sec A. 
These formulse were proved for acute angles in § 4. 

32. Functions of (180° — A) in terms of those of A. 

By § 30, sin (180° -A)= cos (90°-^) = sin A (§ 31). 
cos (180° -A) = - sin (90°-^) = - cos A. 
tan (180° -A) = - cot (90°-^) = - tan A. 
cot (180° -A)=- tan (90°-^) = - cot A. 
sec (180° -A)=- CSC (90°-^) = - sec A. 
CSC (180° -A)= sec (90° -A)= esc A. 



24 



Plane Trigonometry. 



33. By successive applications of the theorem of § 30, 
any function of a multiple of 90°, plus or minus A, may be 
expressed as a function of ^. 

1. Express sin (270° + J.) as a function of A. 

By § 30, sin (270'^+^) = cos (180°+^) = - sin (90°+^) = - cos A. 

If the multiple of 90° is greater than 270°, we may sub- 
tract 360°, or any multiple of 360°, from the angle, in 
accordance with § 19. 

2. Express sec (990° — J.) as a function of A. 
Subtracting twice 360°, or 720°, from the angle, we have 

sec (990°-^) = sec (270°-^). 
And by § 30, sec (270° -A) = - esc (180° - ^) = - esc ^ (§ 32) . 

If the multiple of 90° is negative, we may add 360°, or 
any multiple of 360°, to the angle. 

3. Express tan (— 180° -|- ^) as a function of A. 
Adding 360° to the angle, we have 

tan (- 180° + A') = tan (180° + A). 
And by § 30, tan (180° -hA) = - cot (90° + A') = tan A. 



EXAMPLES. 

Express each of the following as a function of A : 



4. sin (180° + ^). 

5. cos (270°-^). 

6. cot (450° + ^). 

7. CSC (360°-^). 

8. tan (540° - ^). 

9. sec (630° 4-^). 
10. tan (- 270° - ^). 



11. CSC (-90°-^). 

12. cot (- 180° + ^). 

13. sin (- 630° + ^). 

14. tan (- 450° - ^). 

15. cos (- 900° - ^). 

16. sin (810°-^). 

17. CSC (1080° + ^). 



18. sec (1260° + ^). 



Trigonometric Functions. 25 

34. By means of the theorem of § 30, any function of 
any angle, positive or negative, may be expressed as a func- 
tion of a certain acute angle. 

1. Express sin 317° as a function of an acute angle. 

By § 30, sin 317° = cos 227° = - sin 137° = - cos 47°. 

Since the complement of 47° is 43°, another form of the result is 
-sin 43° (§4). 

Note. As in the examples of § 33, 360°, or any multiple of 360°, 
may be added to, or subtracted from, the angle. 

EXAMPLES. 

Express each of the following as a function of an acute 
angle : 

2. cos 322°. 4. sec 559°. 6. cot (- 378°). 

3. tan 208°. 5. esc 803° 45'. 7. sin (- 139° 5'). 

It is evident from the above that any function of any 
angle can be expressed as a function of a certain acute 
angle less than 45°. 

Express each of the following as a function of an acute 
angle less than 45° : 

8. cot 155°. 10. sec 457°. 12. tan (-681°). 

9. sin 1138° 36'. 11. cos 496° 20'. 13. esc (- 257°). 

14. Eind the numerical value of esc (— 210°). 

Adding 360° to the angle, we have 

CSC (-210°)= CSC 150°. 
And by § 30, csc 150° = sec 60° = 2 (§ 8). 

Eind the numerical values of the following : 

15. cot 405°. 17. CSC 600°. 19. cos (-420°). 

16. sin 480°. 18. tan 690°. 20. sec (-225°). 



26 



Plane Trigonometry. 



III. GENERAL FORMULiB. 

35. It follows directly from the definitions of § 17 that, 
if X is any angle, 



sm X 



cos X 



CSC X 

1 



tanic = 

cot X = 



cot it' 
1 



sec X — 



sec X tan x 

36. To iwove the formula 



cscx = 



COS X 

1 

sm a; 



tan a' 



sm a? 

COSiC 



(3) 



(4) 





Jlf 



Pio. 2. 




There may be four cases : x in the first quadrant (Fig. 1), 
X in the second quadrant (Fig. 2), x in the third quadrant 
(Fig. 3), or x in the fourth quadrant (Fig. 4). 

In each case, let the positive angle XOP represent the 
angle x, and draw PM perpendicular to XX'. 



General Formulae. 27 

Then in each figure, by the definitions of § 17, 

ord. P 







tan X = 


ord. P 
abs. P 


dist. P 
abs. P 
dist. P 


sin X 
cos a? 


37. 


Top\ 


rove the 


formula 
Gotx = 


cos a? 
sin .1? 




By§ 


35, 


cot X : 


_ 1 _ 

tan a; 


sm a; ^ sin a.- 



(5) 



cos a.' 

38. To prove the formula 

sin^ X + cos^ x — 1. (6) 

Note. Sin2 aj signifies (sin x)^ ; that is, the square of the sine of x. 

Tliere may be four cases : x in the first quadrant, x in the 
second quadrant, x in the third quadrant, or x in the fourth 
quadrant. 

In each figure of § 36, we have by Geometry, 

pm" + om' = 0P\ 



Dividing by OP , ^^ + -^=- = 1. 

op' op 

But in each figure, 

PM' / ■ s2 a OM' , .2 

^ = (sm xY, and „ = (cos a^)^ ; 

OP op' 

PM PM PM' 

for, whether sin x equals + — - — or , its square is -^ — 

OP OP Qp^ 

Whence, sin^a^ + cos^a^^ 1. 

39. Formula (6) may be written in the forms 

sin^ x = l — cos^ x, and cos^ x = l — sin^ x. 



28 



Plane Trigonometry. 



40. To prove the formulce 

sec^ ic = 1 + tan^ x, 
and ^ csc^ x = 1 -j- cot^ x. 

By (6), 1 = cos^ X + sin^ x. 



(7) 

(8) 

(A) 



Dividing by cos^ x, 



= 1 + 



Whence by (3) and (4), sec^ x = 1 + tan^ x. 
Again, dividing (A) by sin^ x, we have 



1 + 



cos^a? 



snr a? sm" a? 

Whence by (3) and (5), csc^ x = l-[- cot^ x. 

41. To express sin (a? + y) and cos (a? + y) in terms of the 
sines and cosines of x ayid y. 



I. When 


X 


and y are acute 








/ 


/ 


/ 

E 




B 


> 


/ 








A» 















A D 




Fic 


.. 1 








There may be two cases : x-\-y acute (Fig. 1), and x + y 
obtuse (Fig. 2). 

In each figure, let Z DOS = aj and Z BOC = y. 

Then, ZDOC=x-^y. 

From any point (7 in 00 draw lines CA and OjB perpen- 
dicular to OD and 0-B, respectively ; also, draw lines BD 
and BE perpendicular to OD and AC, respectively. 



General Formulae. 29 

Since EC is perpendicular to OD, and BC to OB, angles 
BCE and DOB are equal ; tliat is, Z BCE — x. 
In either figure, by § 17, 

. , , . ^C BD + CE BD , CE 

^BD OB CE BC 
OB OC BC OC 

Then, sin {x + y)= sin a? cos 2/ + cos x sin ?/. (9) 

Again, by § 17, in Fig. 1, cos (aj + y) = -g^, 

OA 

and in Fig. 2, cos (a? + ?/) = - — — . 

Then in either figure, 

OD - BE OD BE 



cos (x-\-y) = 



OC OC OC 



^Op qB_BE BC 
OB OC BC ^ OC 

Then, cos (x -}-y)= cos x cos y — sin x sin y. (10) 

42. Formulae (9) and (10) are very important, and it is 
necessary to prove them for all values of x and y. 

They have already been proved when x and y are any 
two acute angles ; or, what is the same thing, when they 
are any two angles in the first quadrant. 

Now let a and b be any two angles in the first quadrant. 

By § 29, sin [90° + (a + 6)] = cos (a + 6), 
and cos [90° + (a + 6)] = — sin (a -f b). 

Whence, by (9) and (10), 

sin [90° -\- (a-\-b)']= cos a cos b — sin a sin b, (A) 
and cos [90° + (a +&)]=— sin a cos b — cos a sin b. (B) 



30 Plane Trigonometry. 

By § 29, cos a = sin (90°+ a), and — sin a = cos (90°+ a). 

Then, (A) and (B) may be written in the forms 
sin [(90° + a) + 6] = sin (90° + a) cos b + cos (90° + a) sin b, 
cos [(90° + a) + 6] = cos (90° + a) cos b - sin (90° + a) sin b ; 

which are in accordance with (9) and (10). 

But 90° + a is an angle in the second quadrant. 

Therefore, (9) and (10) hold when one of the angles is 
in the second quadrant, and the other in the first. 

In like manner, by supposing a to be any angle in the 
first quadrant, and b any angle in the second, (9) and (10) 
may be proved to hold when both angles are in the second 
quadrant. 

Again, by supposing a and b to be any two angles in the 
second quadrant, (9) and (10) may be proved to hold when 
one angle is in the second quadrant and the other in the 
third; and so on. 

Hence, (9) and (10) hold for any values of x and y what- 
ever, positive or negative. 

43. Putting, in (9) and (10), — y in place of y, 

sin (x — y)= sin a?. cos (~ y)-\- cos a? sin (— y) 

= sin X cos y + cos x(— sin y), by § 2S, 
= sin X cos y — cos x sin y. (11) 

cos (x — y)= cos X cos (— y) — sin x sin (— y) 
= cos X cos y — sin x(— sin y) 
= cos xGosy -\- sin x sin y. (12) 

44. By (4), 

tan(a. + 2/)=^H^^ 
cos {x -\- y) 

= sin X cos 2/ + cos X sin .v^ , . ^^^ .^^. 

COS X COS y — sin x sin y 



General Formulae. 31 

Dividing each, term of tlie fraction by cos .v cos y, 
sin X cos y . cos 2' sin y 

, ^ cos X COS If cos X COS ?/ 

tan (2' + v) = '^^ ^ T-^ 

' COS 2* COS ^ sm .r sm 2/ 

cos X COS ;y cos x cos ?/ 



_ tan X + tan ?/ 
1 — tan X tan ?/ 



In like manner, we may prove 



(13) 



4- / \ tan .r — tan?/ x_^. 

tan (a; — ?/) = -^^- (14) 

^ "^^ l + tan.i'tan?/ ^ ^ 

Again, by (5), cot {,c + ^) = "^^ ^^^' + -^^^ 

sm (x- + y) 

_ cos >!• COS ?/— sin a- sin ?/ 
sin X cos ?/+ cos x sin ?/ 

Dividing each term of the fraction by sin x sin y, 

cos X cos ?/ sin a.' sin y 

, , , sin X sin ?/ sin x sin v 

cot (iC + ?/) = — : : — ^ 

^ sm.rcos^ cos.i'sm^ 

sin X sin y sin x sin ?/ 

_ cot X cot ?/ — 1 
cot y + cot :c 

In like manner, vre may prove 

^/ X C0t.TC0t?/ + l . ^. 

cot ix — y) = '^—^ — (16) 

^ ^^ cot y — cot x ^ ^ 

45. From (9), (10), (11), and (12), we have 

sin (« + h) = sin a cos h + cos a sin h. (A) 

sin (a —1)) = sin o cos 6 — cos a sin 5. (B) 

cos (a + &) = cos a cos 6 — sin a sin 6. (C) 

cos {a— h) = cos a cos 6 + sin a sin 6. (D) 



(15) 



32 Plane Trigonometry. 

Adding and subtracting (A) and (B), and then (C) and (D), 

sin (a + &) + sin (a — b) = 2 sin a cos b. 
sin (a + &) — sin (a — 6) = 2 cos asin b. 
cos (a + 6) 4- cos (a — 6) = 2 cos a cos 6. 
cos (a + 5) — cos (a — 5) = — 2 sin a sin b. 
Let a + 6 = a?, and a — b = y. 
Adding, 2a==x-\-y, or a, = -J- (x + ?/). 

Subtracting, 2b = x — y, or 6 = ^ (ic — ?/). 
Substituting these values, we have 

sin a? + sin 2/ = 2 sm^(x-\-y)GOS^{x — y). (17) 
sin X — sin ?/ = 2 cos -J (a? + 2/) sin i(x~ y). (18) 
cos a? + cos ?/ = 2cos-J-(a^ + 2/) cos-2-(ic — 2/). (19) 
cos a? — cos 2/ = — 2 sin J (cc + 2/) sin ^ (a? — 2/). (20) 

46. By (17) and (18), we have 

sin X + sin y _ 2 sin -J- (a? + 2/) cos ^(x — y) 
. sin ic — sin 2/ ~ 2 cos i (a^ + 2/) sin ^'{x — y) 

= tan|-(a; + y) cot ^(a; - 2/) 

47. Functions of 2 x. 

Putting 2/ = ^ in (9)j we have 

sin 2 i» = sin x cos cc + cos x sin ic 

= 2 sin ic cos a?. (22) 

Putting 2/ = a? in (10), we obtain 

cos 2x = cos a? cos x — sin a? sin a; 

= cos^ X — sin^ X. (23) 



General Formulae. ^^ 

We also have by § 39, 

cos 2 ic = (1 — sin^ x) — sin^ ic = 1 — 2 sin^ x. (24) 

cos 2x = cos^ X— (1 — cos^ 0?) = 2 cos^ x—1. (25) 

Putting 2/ = a? in (13) and (15), 

, o tan X + tan x 2 tan x ,_ _^ 

tan 2 a? = ^ =- ^ — (26) 

1 — tan X tan ic 1 — tan- x 

. o cot 07 cot X — 1 COt^ X — 1 /«„x 

cot 2 a? = = -— (27) 

cot X + cot X 2 cot X 

48. Functions of ^ a?. 

From (24) and (25) we have, by transposition, 

2 sin^ x = l — cos 2 a?, and 2 cos^ a? = 1 + cos 2 a?. 

Putting J X in place of a;, and therefore x in place of 2 a?, 
we have 

2 sin^ ^ aj == 1 — cos x, (28) 

2 cos^ ^ X = 1 + cos X. (29) 

Again, putting -J-x in place of x in (22), 

2 sin ^ a; cos ix = sin x. (A) 

Dividing (28) by (A), 

2 sin^ ^x _ 1 — cos x 



2 sin -J- a? cos -J- a? sin x 

Whence, by (4), tan -i x = ^ ~ ^^^ ^ . (30) 

Dividing (29) by (A), 

2 cos^ ^x _ 1 + cos a; 
2 sin i x cos 4- a? sin x 



cos a; 



Whence, by (5), cot -J- x = iJlidd^. (31) 

sm X 



34 Plane Trigonometry. 

EXERCISES. 

49. 1. Prove the relation sec^ x csc^ x = sec^ x + csc^ 

By (3), sec^x csc^x = --^-^.^ = ^'"'^ + ""f ^ by (6^ 
cos^ X sin2 X cos2 x sm^ x 



cos^ X sin2 X cos2 x sin^ x 
1 , 1 



cos"^ X sin2 X 

sin 5 £c — 

cos 5x -\- cos X 



sec2 X + csc^ X. 



2. Prove the relation = tan 2 x. 

By (18) and (19), 

sin 5 X — sin x _ 2 cos ^(5x + x) sin|(5x — x) _ sin 2 x _ , ^ 
cos5x + cosx 2 cos|(5x + x) cos-|(5x — x) cos2x 

OX) ^1, 14.- tan (.T + v) — tan x . 

3. Prove the relation ^ — -^^ — = tan y. 

1 + tan (x + y) tan x 

By (14). tan(x + i/)-tanx ^ tan [(x + ?/) - x] = tan y. 
•^ ^ ^' 1 +tan(x + 2/)tanx lk ifj j y 

4. Prove the relation sin 3x = Z sin x — 4 sin^ a:;. 

By (18), sin 3x — sin x = 2 cosi(3x + x) sin 1(3 x — x). 

Then, sin 3 x = sin x + 2 cos 2 x sin x 

= sin X + 2 (1 — 2 sin2 x) sin x, by (24) 

= sin X + 2 sin x — 4 sin^ x = 3 sin x — 4 sin^ x. 

The artifice used above is advantageous in finding the 
sine or cosine of any odd multiple of x. 

. Prove the following relations : 

K sin {x + y) __ tan x + tan y 
sin (x — y) tan x — tan y 

^ cos (x + I/) _ cot X cot ?/ — 1 
cos (x — y) cot a? cot 2/ + 1 

i~ cos a? + cos y ^ i / , \ j. i / \ 

7. -^^ — ^ = — cot i (it- + y) cot i (a^ — 2/)- 

cos X — cos 2/ 



General Formulae. 35 



8. sm3x + sinx^^^^^^ 
COS 3x -\- cos X 

9. sec^ A + tan^ 5 = sec^ B + tan^ A. 

cos a? cos y 

t^n(A + B) + t^n(A~B) ^^^^^^^ 
1 - tan (A + 5) tan (A - 5) 

. 12. cos X cos (a? — 2/) + sin x sin (x — y) = cos 2/. 

^13. sin 5 aj cos 2x — cos 5 a? sin 2 x = sin 3 ic. 

- M sin 3 a? — sin 5 .t , . 

14. = — cot 4 a;. 

cos 3x — cos 5 ic 

1 c cot X + tan a? o 

15. — = sec2aj. 

cot X — tan X 

16. cos 3x = 4: cos^ X — 3 cos ic. 

17. tan^ia^^ ^~^^^^ > 

1 + cos a^ 

18. sin (x-\-y -{- z) = sin .t cos ?/ cos 2 + cos x sin ?/ cos 2 

H- cos a? cos ?/ sin z — sin a.- sin y sin 2. 

19. cos (x -]- y + z) = cos a^ cos y cos 2; — sin x sin ?/ cos ;2 

— sin X cos ?/ sin z — cos .^' sin y sin 2. 

rtfk ^ / , , \ tanflj+tanw+tans; — tancc tan?/ tanz 

20. tdi,ii(x-{-y^z) = - ^-^ ^ 

1 — tan 37 tan ?/— tan 2/ tan 2; — tan 2; tan a; 

21. Prove the relation of Ex. 4 by putting x = 2x, and 

y = x, in (9). 

Prove the relations : 

nn . o 3 tan X — tan^ x 

22. tan3a^==— 

1-3 tan2 X 

23. sin (x + y) sin (x — y)— sin^ a? — sin^ i/- 

24. cos (a; + i/) cos (x — y) = cos^ a) — siu^ y. 



^6 Plane Trigonometry, 

2 tana? 



25. sin 2 x = 

26. cos2x = 



1 + tan^ X 
1 — tan^ X 



1 + tan^ X ^ 

27. 4 cos X cos (60° + x) cos (60° —x) = cos 3 x. 

28. tan (45° + x) - tan (45° - ic) = 2 tan 2 x. 

29. cos^ (x-i-y)— sin^ a? = cos (2 x-\-y) cos 2/. 

30. cot^- cot 2^ = CSC 2 A 

31. sin4x + sin_3x^^^^^^^ 
cos 3 X — COS 4 X 

oo sin 50° - sin 10° ^ /p - - 

' cos 50° -cos 10° * 

33. cos 80° + cos 40° = cos 20°. 

34. sin 4:X = 4:smx cos x—S sin^ x cos x. 

35. cos 4 a; = 1 — 8 cos^ x-\-S cos^ a?. 

36. cos 5x = 5 cos a? — 20 cos^ x + 16 cos^ x. 

37. cos (2 X + ?/) + 2 sin aj sin (x-{-y) = cos ?/. 

38. By putting x = 45° and y = 30° in (11) and (12), prove 

sin 15°= ^-^^ , cos 15°==^ + A 
4 4 

39. By putting x = 30° in (30) and (31), prove 

tan 15° = 2 -V3, cotl5° = 2+V3. 

40. By putting x = 15° in (3), and using the results of 
Ex. 38, prove 

secl5°=V6-V2, cscl5° = V6 + V2. 

41. By putting x = 45° in (28) and (29), prove 

sin 221° = JV2-V2, cos 22i-° = iV2 + a/2. 



General Formulae. 37 

42. By putting x = 45° in (30) and (31), prove 

tan 22^° = V2 - 1, cot 22^° = V2 + 1. 

43. By putting x = 22j-° in (7) and (8), and using tlie 
results of Ex. 42, prove 



sec22i° = V4-2V2, esc 22^° = V 4 -f- 2 V2. 

Prove the relations : 

44. cos^ X — sin* x = cos 2 x. 

. e CSC X + cot X ,0 1 

45. — = cot^ i X. 

CSC X — cot X 

46. sin (90° + A) -\- sin (210° + ^) + sin (210° - ^) = 0. 

47. (sin X + sin yY + (cos a? + cos y^ = 4 cos^ ^ ~" ^ • 

48. tan 2 a? cot a^ — 1 = sec 2 a?. 

49. tan x tan (60° + x) tan (120° + x) = tan 3 x. 

KQ 1 + sin 2 X _ co s X + sin .t 

cos 2 a? cos a^ — sin x 

4 tan a^ — 4 tan^ x 



51. tan 4 a? = 



1 — 6 tan^ X -}- tan* aj 



52. ^ ''^^ ^tan(45°-ix). 

cosx ^ 

., 2 tan (45° -a^) ^ 

53. ^ , ^ ' ..,0 — ^ = cos 2 a?. 
1 + tan^ (45 — x) 

54. Prove the first result of Ex. 43 by putting x = 221' 
in (3), and using the result of Ex. 41. 



38 Plane Trigonometry. 



IV. MISCELLANEOUS THEOREMS. 

50. Circular Measure of an Angle. 

An angle is measured by finding its ratio to another angle, 
adopted arbitrarily as the unit of measure. 

The usual unit of measure for angles is the degree, which 
is an angle equal to the ninetieth part of a right angle. 

Another method of measuring angles, and one of great 
importance, is known as the Circular Method ; in which the 
unit of measure is the angle at the centre of a circle subtended 
by an arc ivhose length is equal to the radius. 

Note. The unit of circular measure is called a radian. 




Thus, let AOB be any central angle, and AOC the unit 
of circular measure ; that is, the angle at the centre sub- 
tended by an arc whose length is equal to OA. 

AOB 



Then, circular measure AOB 
But by Geometry, 



.AOC 
ZAOB arc .45 arc ^B 



Whence, circular measure A OB 



AOC diVcAC OA 
arc AB 



OA 



That is, the circular measure of an angle is the ratio of its 
subtending arc to the radius of the circle. 

51. By § 50, the circular measure of a right angle is the 
ratio of one-fourth the circumference to the radius. 

But if R denotes the radius, the circumference is 2 irR, 



Miscellaneous Theorems. 39 

Whence, circular measure 90° = ^ — _-!I — = ^. 

R 2 

It follows from the above that the circular measure of 
180° is TT ; of 60°, | ; of 45°, ^ ; etc. 

That is, an angle expressed in degrees may he reduced to 
circular measure by finding its ratio to 180°, and midtiplying 

the result by ir. 

23 
Thus, since 115° is — of 180°, the circular measure of 
90 36 

115° is ^^. 
36 

52. Conversely, an angle expressed in circular measure 
may be reduced to degrees by 'tnultiijlying by 180° and dividing 
by IT ; or, more briefly, by substituting 180° for tt. 

Thus, I^ = X of 180° = 84°. 



15 15 



53. In the circular method, such expressions may occur 
as "the angle |," "the angle 1," etc. 

These refer to the unit of circular measure; thus, the 
angle |- signifies an angle whose subtending arc is two-thirds 
of the radius. 

The angle 1 signifies the angle whose subtending arc is 
equal to the radius, or the unit of circulai measure. 

The angle 1, reduced to degrees by the first rule of § 52, 
gives 

^^= o-.!f^I — =57.2958°, approximately. 
TT 3.14159 ••• 

Then the rule of § 52 may be modified as follows : 

An angle expressed in circular measure may be reduced to 
degrees by multiplying by 57.2958°. 

Thus, the angle f 

= I X 57.2958° = 38.1972° = 38° 11' 49.92". 



40 Plane Trigonometry. 



EXAMPLES. 

54. Express each, of the following in circular measure : 

1. 120°. 3. 67° 30'. 5. 86° 24'. 7. 163° 7' 30". 

2. 315°. 4. 146° 15'. 6. 53° 20'. 8. 88° 53' 20". 

Express each of the following in degree measure : 

9. ^. 11. li^. 13. 5. 15. ^. 

6 81 2 3 

10. y^. 12. ??^. 14. i. 16. -' 

24 64 4 5 

55. Inverse Trigonometric Functions. 

The expression sin~^cc, called the inverse sine of x, or the 
anti-sine of x, signifies the angle ivhose sine is x. 

Thus, the statement that the sine of the angle x is equal 
to y may be expressed in either of the ways 

sin x = y, or a? = sin~-^ y. 

In like manner, cos~^ x signifies the angle whose cosine is 
X ; tan~^ X, the angle whose tangent is x ; etc. 

Note. The student must be careful not to confuse the above nota- 
tion with the exponent — 1 ; the — 1 power of sin x is expressed 

(sin x)"i, and not sin-i x. 

It is evident that the sine of the angle whose sine is x is 
X ; that is, sin (sin~^ x) = x. 

In like manner, cos (cos"^ x) =: x-, tan (tan~^ aj) = x ; etc. 

56. By aid of the principles of § 55, we may derive from 
any formula involving direct functions a relation between 
inverse functions. 

1. From the formula t2iii(x + y)= tana; + tan?/ ^ ^^^^^ 

1 — tan X tan v 



.tan-i a + tan-^ b = tan-^ -^±A„ 
1 — ab 



Miscellaneous Theorems. 

Let tan x = a, and tan y = b. 

Then by § 55, x = tan-i a, and y = tan-i b. 
Substituting these values in the given formula, 

a + b 



41 



Whence, 



tan(tan-i a + tan~^ 6) = 



tan-i a + tan-i b = tan" 



1 -ab 



1 -ab 



2. Prove the relation cot~^a— sec"^ & = cos 

Let cot-i a = x, and sec-i b =y. 

Then, cot x = a, and sec y = b. 

Now,, cos(x - y)= cos X cos j/ + sin x sin y. (A) 

To find the sines and cosines of x and y, we use the method of § 6. 





In the right triangle containing angle x, the adjacent side is a, and 
the opposite side 1 ; then, the hypotenuse is Va^ + 1. 

In the right triangle containing angle y, the hypotenuse is b, and 
the adjacent side 1 ; then, the opposite side is Vft^ _ i. 

Substituting the values of cosx, cosy, sin x, and siny in (A), we 
have 

a 1 



cos(x — y) 



+ 



V^^'+l ^ Va2 + 1 



^ & Va2 + 1 



Whence, x — y or cot"i a — sec-i & = cos-i 



V¥ 



EXAMPLES. 



3. From cot 2x = 



cot^a.'— 1 



bVa^Tl 



^ n — nr»f ~1 



prove 2cot~^a=cot 



2 cot i» 
4. From cos 2 a? = 1 — 2 sin^ a;, prove 

2sin-ia = cos-\l-2a2). 



2a 



42 Plane Trigonometry. 

5. From sin 2x = 2 sin x cos x^ prove 



2 cos^^ a = sin~^ (2 aV 1 — a^). 
% 6. From cos (a? + ?/) = cos x cos y — sin a? sin ?/, prove 
cos~^ a + cos~^ 6 = cos~^(a?> — Vl — a^Vl — 6^). 

Prove the following relations : 
^7. cot-i a -h cot-i h = cot-i ^^^. 
8. 2cos-ia = cos-\2a2-l). 



9. sin~^ a — sin~^ & — sin~^ (a Vl — h^— 5 Vl — a^). 

10. 3 sin-i a = sin-^ (3 a - 4 a^). (Ex. 4, p. 34.)' 

11. tan-^ a + cot-^ 6 = tan-^ ^^5_±1. 

6 — a 

12. cot-^a -h)- cot-i (a + 6) = cot-^ a^-^^ + l . 



26 
■^13. tan~^a = sin~^ — » 



^ , . , , Va^ — 1 
-' 14. csc~^ a — cos~^ 



15. sin- a + cos- & = tan- a^> + Vl - aVl : ^. 

5Vr^^^-aVl-&^ 



- « 1 ] , 1 Va- — 1 + V6^ — 1 

#16. sec— a — csc-^ 6 = cos— -^^ ^^^-^ 

ah 



1 K- , 1 , _i 1 ■ 1 a + Va^ — 1 
17. tan~^a + cos— -^sm"^ 

« a-^d^ + 1 



18. 2 sin- a = tan- ^^^%^'. 

1 — 2a^ 

19. tan-— ^-tan-^-^^i^ = tan--i- 

a — 1 a 2 a 



20. 2 sec"^ a = cot 



Miscellaneous Theorems. 



43 



2 Va^ - 1 



57. The following table expresses the value of each of 
the six principal functions in terms of the other five : 



sin 

cos 
tan 
cot 
sec 

CSC 






tan 


1 




J_ 

CSC 


Vsec2 - 


1 


Vl — cos^ 






sec 




Vl -f tan^ 
1 


Vl + cof- 


cot 1 


VCSC2 - 1 


Vl - sin2 
sin 




— 






CSC 

1 


Vl + tan" 

1 
tan 


Vl + cof^ 

1 
cot 


sec 


Vl - cos- 


Vsec2 - 
1 


1 


Vl — sin^ 


cos 
cos 


VcSC2 — 1 


Vl — sin^ 


VCSC2 - 1 
CSC 


sin 
1 






Vl — cos^ 

cos 
1 


Vsec2 - 
sec 


1 


Vl + C0t2 


Vl + tan2 


Vl - sin2 

1 

sin 


cot 


VCSC2 - 1 


a/1 _l t,P,n2 




Vl + C0t2 


Vl — C0S2 


tan 


Vsec2 - 


^ 



The reciprocal forms were proved in § 35. 
The others may be derived by aid of §§ 35, 36, 37, 39, 
and 40, and are left as exercises for the pupil. 
As an illustration, we will prove the formula 



cos^ = 



By § 39, 



Vcsc^ A - 
csc^ 



COS A — V 1 — sin^ A 



4 



VcscM-1 



csc^^ 



csc^ 



They may also be conveniently proved by the method of 
§ 6 ; thus, let it be required to prove the formula for each 
of the other functions in terms of the secant. 



We have 



sec^ = 



sec J. 



44 



Plane Trigonometry. 
B 




Since the secant is the ratio of the hypotenuse to the 
adjacent side, we take AB=s,qq,A, and AC = 1. 



Whence, 50 = V^s' -^O' = VsecM -1. 

Then by definition, 



. Vsec^ A — 1 
sm A = — ) 



sec^ 



cos A = -J 

sec J. 



cot^ 



CSC J. = 



VsecM-1 

sec A^ 
VsecM-1 



tan A = V sec^ A — 1, 



58. Line Values of the Functions. 



Let XOB be any angle. 

With as a centre, and a radius equal to 1, describe cir- 
cumference AB, cutting OX at A, OB at B, and F at C. 





Miscellaneous Theorems. 



45 









F 






X 








y 


C 




■r> 


-»-r 




C 








■^ 


y 


.Fvr 


y-^ 


■ 


\ 




/ 


^ 


-. > 


r 




< ^ 


^ 


\ 


'i 


P 


( 


^ 1 




V' / 


IK 





^i> 


A , 




I 


>» 


K 


° J 


A ^ 


I 


V 


\ 


. 1 






\ 
B 






Ly 






y. V^ 




^ 


^ 






] 


r' 






F' 










Fia 


.3. 






Fig 


.4. 







Draw line BD perpendicular to XX' ; also, lines AE and 
CF perpendicular to OX and Y, respectively, meeting OB, 
or OB produced, at E and F, respectively. 

Then by § 17, the functions of Z XOB are : 





Sin. 


Cos. 


Tan. 


Cot. 


Sec. 


Csc. 


Fig. 1. 


BD 
OB 


OD 
OB 


Bn 

on 


OD 
BD 


OB 
OD 


OB 
BD 


Fig. 2. 


BD 
OB 


OD 
OB 


BD 
OD 


OD 
BD 


OB 

OD 


OB 
BD 


Fig. 3. 


BD 
OB 


OD 
OB 


BD 
OD 


OD 
BD 


OB 
OD 


OB 
BD 


Fig. 4. 


BD 
OB 


OD 
OB 


BD 
OD 


OD 
BD 


OB 
OD 


OB 
BD 



Now right triangles OBD, OEA, and OCF are similar, 
since their sides are parallel each to each. 
Then, since OA = OC = 1, we have 



?D^AE^^ 
OD OA 

2^ = ^=CF 
BD 00 



OB^OE 
OD OA 

OB^OF 
BD OC 



= OE, 



OF. 



4.6 Plane Trigonometry. 

Whence, since OB = 1, the functions of Z XOB are : 





Sin. 


Cos. 


Tan. 


Cot. 


Sec. 


Csc. 


Fig. 1. 


BD 


OD 


AE 


CF 


OE 


OF 


Fig. 2. 


BD 


-OD 


-AE 


-CF 


-OE 


OF 


Fig. 3. 


-BD 


-OD 


AE 


CF 


-OE 


-OF 


Fig. 4. 


-BD 


OD 


-AE 


-CF 


OE 


-OF 



That is, iftJie radius of the circle is 1, 

The sine is the perpendicular drawn to XX' from the 
intersection of tire circumference with the terminal line. 

The cosine is the line drawn from the centre to the foot of 
the sine. 

The tangent is that portion of the geometrical tangent to 
the circle at the intersection of its circumference with OX 
included between OX and the terminal line, produced if 
necessary. 

The cotangent is that portion of the geometrical tangent 
to the circle at the intersection of its circumference with 
OY included between OY and the terminal line, produced 
if necessary. 

The secant is that portion of the terminal line, or terminal 
line produced, included between the centre and the tangent. 

The cosecant is that portion of the terminal line, or termi- 
nal line produced, included between the centre and the 
cotangent. 

And with regard to algebraic signs. 

Sines and tangents measured above XX' are positive, and 
below, negative ; cosines and cotangents measured to the right 
of YY' are positive, and to the left, negative; secants and 
cosecants measured on the terminal line itself are positive, 
and on the terminal \mQ produced through 0, negative. 

The above are called the line values of the trigonometric 
functions. 



Miscellaneous Theorems. 



47 



They simply represent the values of the functions when 
the radius is 1 ; that is, the mimerical value of the sine of an 
angle is the same as the number which expresses the length 
of the perpendicular drawn to XX' from the intersection of 
the circumference and terminal line. 



59. To trace the changes in the sine, cosine, tangent, cotan- 
gent, secant, and cosecant of an angle as the angle increases 
from 0° to 360". 

^4 -F3 c| Foj 




Let AB^ be a circle whose radius is 1. 

Let the terminal line start from the position OA, and re- 
volve about point as a pivot towards the position 0(7. 

Then since the sine of the angle commences with the 
value 0, and assumes in succession the values B^D^, B2D2, 
OC, B^Dg, BJ)^, etc. (§ 58), it is evident that, as the angle 
increases from 0° to 90°, the sine increases from to 1 ; 
from 90° to 180°, it decreases from 1 to ; from 180° to 270°, 
it decreases (algebraically) from to — 1 ; and from 270° to 
360°, it increases from — 1 to 0. 

Since the cosine commences with the value OA, and 
assumes in succession the values OD^, OD^, 0, — OD^, — OD^, 
etc., from 0° to 90°, it decreases from 1 to ; from 90° to 
180°, it decreases from to - 1 ; from 180° to 270°, it in- 
creases from — 1 to ; and from 270° to 360°, it increases 
from to 1. 



48 Plane Trigonometry. 

Since the tangent commences with the value 0, and as- 
sumes in succession the values AEi, AE^, 00 (see Note to 
§ 25), -AE^, -AE^, etc., from 0° to 90°, it increases from 
to 00 ; from 90° to 180°, it increases from — 00 to ; from 
180° to 270°, it increases from to 00 ; and from 270° to 
360°, it increases from — 00 to 0. 

Since the cotangent commences at 00, and assumes in suc- 
cession the values CF^, CF^, 0, — GF^, — CF^, etc., from 0° 
to 90°, it decreases from 00 to ; from 90° to 180°, it de- 
creases from to — 00 ; from 180° to 270°, it decreases from 
00 to ; and from 270° to 360°, it decreases from to — 00. 

Since the secant commences with the value OA, and as- 
sumes in succession the values OEi, OE2, 00, —OE^, —OE^, 
etc., from 0° to 90°, it increases from 1 to 00 ; from 90° to 
180°, it increases from - 00 to - 1 ; from 180° to 270°, it 
decreases from — 1 to — 00 ; and from 270° to 360°, it de- 
creases from 00 to 1. 

Since the cosecant commences at 00, and assumes in suc- 
cession the values OF^, OF^, OC, OF^, OF^, etc., from 0° to 
90°, it decreases from 00 to 1 ; from 90° to 180°, it increases 
from 1 to 00 ; from 180° to 270°, it increases from — 00 to 
— 1 ; and from 270° to 360°, it decreases from — 1 to — 00. 

__ — . ... vr 1 p sm i^ , tans? 
60. Limiting Values of and 

r^ ^ .-,-,.. . -, ^7/. . sin X , tan x 
To find the limiting values of the fractions ana 

when X is indefinitely decreased. 

Note. We suppose x to be expressed in circular measure (§ 50). 

P 




Miscellaneous Theorems. 49 

Let OPXP^ be a sector of a circle ; Z POP being < 180°. 
Draw lines PT and PT tangent to the arc at P and P\ 
respectively; also, lines OT and PP^ intersecting at M. 

By Geometry, PT = P T. 

Then, OT bisects PP at right angles, and also bisects 
arc PP^ at X. 

Let Z XOP = Z XOP' = X. 

By Geometry, arc PP^ > chord PP', and < PTP. 

Whence, arc PX>PJf, and <PT 

^. . SiTcPX^PM ^ ^PT 

Therefore, ___>_, and <— • 

Or by § 50, circ. meas. x > sin x, and < tan x. 

Eepresenting the circular measure of ic by cc simply, and 
dividing through by sin a?, we have 

^ >1, and <*?^ or -i- (§ 36). 



sm X sm X cos x 



-rxn smx . , 
Whence, <1, and >cosic. 

' X 



But when x is indefinitely decreased, cos x approaches the 
limit 1 (§ 22). 

Hence, approaches the limit 1 when x is indefinitely 

X 

decreased. 

tana? sin a? since 1 

Again, = — — = X 

° ' X a? cos a? x cosic 

sm X 1 

But and approach the limit 1 when x is indefi- 

X cos a? 

nitely decreased. 

Hence, approaches the limit 1 when x is indefinitely 

X 

decreased. 



50 Plane Trigonometry. 



V. LOGARITHMS. 

61. Every positive number may be expressed, exactly or 
approximately, as a power of 10. 

Thus, 100 = 102; 13 = 10^-1139 ■-. e|-(,_ 

When thus expressed, the corresponding exponent is called 
its Logarithm to the Base 10. 

Thus, 2 is the logarithm of 100 to the base 10 5 a relation 
which is written logio 100 = 2, or simply log 100 = 2. 

62. Logarithms of numbers to the base 10 are called 
Common Logarithms, and, collectively, form the Common 
System. 

They are the only ones used for numerical computations. 

Any positive number, except unity, may be taken as the 
base of a system of logarithms ; thus, if w" — m, where a 
and m are positive numbers, then x = log„ m. 

Note. A negative number is not considered as having a logarithm. 



63. We have 


by Algebra, 




10^ = 1, 




io-' = i = .i, 


10^ = 10, 




io- = ii-.= -oi. 


10^ = 100, 




10-^ = 4= -001, etc. 



10^ 
Whence by the definition of § 61, 

log 1 = 0, log .1 = - 1 = 9 - 10, 

log 10 = 1, log .01 = - 2 = 8 - 10, 

log 100 = 2, log .001 = - 3 = 7 - 10, etc. 

Note. The second form for log. 1, log .01, etc., is preferable in 
practice. If no base is expressed, the base 10 is understood. 



Logarithms. 51 

64. It is evident from § 63 that the logarithm of a num- 
ber greater than 1 is positive, and the logarithm of a num- 
ber between and 1 negative. 

65. If a number is not an exact power of 10, its common 
logarithm can only be expressed approximately. 

The integral part of the logarithm is called the character- 
istic, and the decimal part the mantissa. 

For example, log 13 = 1.1139. 

Here, the characteristic is 1, and the mantissa .1139. 

For reasons which will appear hereafter, only the man- 
tissa of the logarithm is given in a table of logarithms of 
numbers ; the characteristic must be found by aid of the 
rules of §§ Q)^ and 67. 

66. It is evident from § 63 that the logarithm of a num- 
ber between 

1 and 10 is + a decimal ; 

10 and 100 is 1 + a decimal ; 

100 and 1000 is 2 + a decimal ; etc. 

Therefore, the characteristic of the logarithm of a number 
with one place to the left of the decimal point, is ; with 
two places to the left of the decimal point, is 1 ; with three 
places to the left of the decimal point, is 2 ; etc. 

Hence, the characteristic of the logarithm of a number 
greater than 1 is 1 less than the number of places to the left of 
the decimal point. 

For example, the characteristic of log 906328.5 is 5. 

67. In like manner, the logarithm of a number between 

1 and .1 is 9 + a decimal — 10 ; 
.1 and .01 is 8 H- a decimal — 10 ; 
.01 and .001 is 7 + a decimal — 10 5 etc. 



52 Plane Trigonometry. 

Therefore, the characteristic of the logarithm of a decimal 
with no ciphers between the decimal point and first signifi- 
cant figure, is 9, with — 10 after the mantissa ; of a decimal 
with one cipher between the point and first significant figure 
is 8, with — 10 after the mantissa ; of a decimal with two 
ciphers between the point and first significant figure is 7, 
with — 10 after the mantissa ; etc. 

Hence, to find the characteristic of the logarithm of a num- 
ber between a7id 1, subtract the number of ciphers between 
the decimal point and first significant figure from 9, writing 

— 10 . after the mantissa. 

For example, the characteristic of log .007023 is 7, with 

— 10 written after the mantissa. 

Note 1. It is customary in practice to omit the — 10 after the man- 
tissa of a negative logarithm ; but it should he allowed for in the result. 

Beginners should always write it. 

Note 2. Some writers combine the two portions of the character- 
istic, and write the result as a negative characteristic before the 
mantissa. 

Thus, instead of 7.6036 - 10, the student will frequently find 3.6036, 
a minus sign being written over the characteristic to denote that it 
alone is negative, the mantissa being always positive. 

PROPERTIES OF LOGARITHMS. 

68. In any system, the logarithm of 1 is 0. 

For by Algebra, a*' = 1 ; whence by § ^2, log„ 1 = 0. 

69. /n any system, the logarithm of the base is 1. 
For a} = a', whence, log„ a = 1. 

70. In any system whose base is greater than 1, the logor 
rithm ofOis — oo. 

For if a is greater than 1, a""^ = — = — = 0. 

Whence by § 62, log„ = — oo. 

Note. No literal meaning can be attached to such a result as 

lOga0=-0O. 



Logarithms. 53 

It must be interpreted as follows : 

If, in any system whose base is greater than unity, a number ap- 
proaches the limit 0, its logarithm is negative, and increases without 
limit in absolute value. 

71. In any system, the logarithm of a product is equal to 
the sum of the logarithms of its factors. 

Assume the equations 



«" = ™ I ; whence by § 62, | "= = ]°S' «' 
a^ = n ) (y = log^ n. 

Multiplying the assumed equations, 

a"" X a^ = mn, or a^+^ = mn. 

Whence, log« mn = a? + 2/ = log^ m + log^ n. 

In like manner, the theorem may be proved for the prod- 
uct of three or more factors. 

72. By aid of § 71, the logarithm of a composite number 
may be found when the logarithms of its factors are known. 

1. Given log2 = .3010 and log3 = .4771 ; find log 72. 

log 72 = log(2 X2x2x3x3) 

= log2+ log2 + log2 + logs + logs (§ 71) 

= 3 X log 2 + 2 X log 3 = .90S0 + .9542 = 1.8572. 

EXAMPLES. 

Given log 2 = .3010, log 3 = .4771, log 5 = .6990, and 
log 7 = .8451, find: 

2. log 35. 6. log 147. 10. log 288. 14. log 2205. 

3. log 30. 7. log 225. 11. log 686. 15. log 7875. 

4. log 98. 8. log 175. 12. log 504. 16. log 5832. 

5. log 84. 9. log 420. 13. log 375. 17. log 14112. 



54 Plane Trigonometry. 

73. In any system, the logarithm of a fraction is equal to 
the logarithm of the num.erator minus the logarithm of the 
denominator. 

Assume the equations 

«" = ™l; whence, I ^■ = l°g«'«' 
a^ = n ) (y = log„ n. 

Dividing the assumed equations, 

a^ m , „ m 

— = —, or a*-^ = — 
a^ n n 

m 
Whence, loga— =x — y =^ log,^ m — log^ n. 

74. 1. Given log 2 = .3010; find log 5. 

log 5 = log — = log 10 - log 2 (§ 73) = 1 - .3010 = .6990. 



EXAMPLES. 

Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find : 

2. log^. 5. log33i. 8. log^. 11. log23f 

3. log I. 6. logg. 9. log4f 12. log^. 

4. log 45. 7. log 105. 10. log 525. 13. log96f. 

75. In any system, the logarithm of any power of a quantity 
is equal to the logarithm of the quantity multiplied by the ex- 
ponent of the power. 

Assume the equation a"" = m ; whence, x = log„ m. 

Raising both members of the assumed equation to thepth 
power, 

gpx _ r^p . wixence, log„?/i.^ = px = p log„ m, 



Logarithms. 55 

76. In any system, the logarithm of any root of a quantity 
is equal to the logarithm of the quantity divided by the index 
of the root. 

r — - 1 

For, logaTs/m = log„(»r) = - log„ m (§ 75). 

77. 1. Given log 2 = .3010; find log 2^ 

log 2 3 = - X log 2 = - X .3010 = . 5017. 
^ 3 ° 3 

Note. To multiply a logarithm by a fraction, multiply first by 
the numerator, and divide the result by the denominator. 

2. Given log 3 = A771 ; find log ^3. 

log^ = 1^ = 1^111 = .0596. 
8 8 

EXAMPLES. 
Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find : 

3. log2^ 6. log42^ 9. log -^3. 12. log ^28. 

4. log7l 7. logl5l 10. log ^7. 13. log ^324. 

5. log 5k 8. log48t 11. log ^5. 14. log -^735. 
15. Find log (2* x3*). 

By § 71, log (23 X 3*)= log23 + log 3* = ilog2 + f log3 
= .1003 + .5964= .6967. 

Find the values of the following : 

16. log^i. 18. log||. 20. log:|p. 23. log(Mj*. 

5 5. 

17. log5A/2. 19.log-. 21. log ^. 23. log (23 X 2lK 

5^ ■ V7 



56 Plane Trigonometry. 

78. To prove the relation 

, log„ m 

logj m = , ° , « 
log„6 

Assume the equations 

w" = m) ^ {x = log„ m, 

V ; whence, ^ 
b^ = m) (y = i-Ogj, m. 

From the assumed equations, 

a"" = ¥. 
Taking the ytli root of both members, 

a^ = b. 

Therefore, log„ 6 = -, or y = 



y' iog> 

That is, logft m = , . 

log„& 

79. jTo ^rove ^/ie relation 

logs a X log„ h = l. 
Putting m = a in the result of § 78, we have 

logft a = ^-^^ == . (§ 69). 

log„6 log„6 ^ ^ 

Whence, logj a x log^ 6 = 1. 

80. In the common system., the mantissce of the logarithms 
of numbers having the same sequence of figures are equal. 

Suppose, for example, that log 3.053 = .4847. 

Then, log 305.3 = log (100 x 3.053) = log 100 + log 3.053 
= 2 + .4847 = 2.4847 ; 

log .03053 = log (.01 X 3.053) = log .01 + log 3.053 
= 8 - 10 + .4847 = 8.4847 - 10 ; etc. 



Logarithms. 57 

It is evident from the above that, if a number be multi- 
plied or divided by any integral power of 10, producing 
another number with the same sequence of figures, the man- 
tissse of their logarithms will be equal. 

The reason will now be seen for the statement made in 
§ 65, that only the mantissas are given in a table of loga- 
rithms of numbers. 

For, to find the logarithm of any number, we have only 
to take from the table the mantissa corresponding to its 
sequence of figures, and the characteristic may then be 
prefixed in accordance with the rules of §§ 66 or 67. 

Thus, if log 3.053 = .4847, then 

log 30.53 = 1.4847, log .3053 = 9.4847 - 10, 

log 305.3 = 2.4847, log .03053 = 8.4847 - 10, 

log 3053. = 3.4847, log .003053 = 7.4847 - 10, etc. 

This property is only enjoyed by the common system of 
logarithms, and constitutes its superiority over others for 
the purposes of numerical computation. 

81. 1. Given log 2 = .3010, log 3 = .4771 ; find log .00432. 
We have log 432 = log (2* x 3^) = 4 log 2 + 3 log 3 = 2.6353. 
Then by § 80, the mantissa of the result is .6353. 
Whence by § 67, log .00432 = 7.6353 - 10. 

EXAMPLES. 

Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find : 

2. log 2.4. 6. log .00135. 10. Jog .1029. 

3. log 16.8. 7. log 5880. 11. log 201.6. 

4. log .81. 8. log .0245. 12. log ^7:5. 
6. log .0192. 9. log .000486. 13. log (12.6) t- 



58 Plane Trigonometry. 

USE OF THE TABLE OF LOGARITHMS OF NUMBERS. 

(For directions as to the use of the Table of Logarithms 
of Numbers, see pages 1 to 4 of the Introduction to the 
author's New Four Place Logarithmic Tables.) 

EXAMPLES. 

82. Find the logarithms of the following numbers ; 



1. 


80. 


6. 


.03294. 


11. 


.0007178. 


2 


6.3. 


7. 


.5205. 


12. 


5.1809. 


3. 


.298. 


8. 


20.08. 


13. 


1036.5. 


4. 


772.3. 


9. 


92461. 


14. 


.086676. 


5. 


1056. 


10. 


.0040322. 


15. 


.000011507. 



Find the numbers corresponding to the following loga- 
rithms : 

16. 1.8055. 21. 8.1646-10. 26. 1.6482. 

17. 9.4487-10. 22. 7.5209-10. 27. 6.0450-10. 

18. 0.2165. 23. 2.0095. 28. 4.8016. 

19. 3.9487. 24. 0.9774. 29. 8.1142-10. 

20. 2.7371. 25. 9.3178-10. 30. 5.7015-10. 

APPLICATIONS. 

83. The approximate value of an arithmetical quantity, 
in which the operations indicated involve only multiplica- 
tion, division, involution, or evolution, may be conveniently 
found by logarithms. 

The utility of the process consists in the fact that addi- 
tion takes the place of multiplication, subtraction of divi- 
sion, multiplication of involution, and division of evolution. 

Note. In computations with four-place logarithms, the results 
cannot usually be depended upon to more than four significant figures. 



Logarithms. 59 

84. 1. Find the value of .0631 x 7.208 x .51272. 
By § 71, log (.0631 x 7.208 x .51272) 

= log .0631 f log 7.208 + log .51272. 

log .0631 = 8.8000-10 
log 7.208 = 0.8578 
log. 51272= 9.7099-10 
Adding, log of result = 19.3677 - 20 

= 9.3677 - 10. (See Note 1.) 

Number corresponding to 9.3677 - 10 = .2332. 

Note 1. If the sum is a negative logarithm, it should be written 
in such a form that the negative portion of the characteristic may be 
-10. 

Thus, 19.3677 - 20 is written in the form 9.3677 - 10. 

2. Find the value of ??^. 

7984 

By § 73, log f 1^ = log 336.8 - log 7984. 

log 336. 8 = 12. 5273 - 10 (See Note 2. ) 
log 7984 = 3.9022 
Subtracting, log of result = 8.6251 — 10 

Number corresponding = .04218. 

Note 2. To subtract a greater logarithm from a less, or to subtract 
a negative logarithm from a positive, increase the characteristic of the 
minuend by 10, writing — 10 after the mantissa to compensate. 

Thus, to subtract 3.9022 from 2.5273, write the minuend in the form 
12.5273 - 10 ; subtracting 3.9022 from this, the result is 8.6251 - 10. 

3. Find the value of (.07396)^. 

By § 75, log (.07396)5 = 5 x log .07396. 



log 


.07396 


= 8.8690- 


-10 
5 








44.3450 - 


-50 








= 4.3450- 


- 10 (See 


Note 1, 


•> 






= log. 000002213. 







6o Plane Trigonometry. 



4. Find the value of V.035063. 



By § 76, log V.035063 = i log .035063. 

log .035063 = 8.5449 -10 

20. - 20 (See Note 3.) 

3 )28.5449 - 30 

9.5150 - 10 = log .3274. 

Note 3. To divide a negative logarithm, write it in such a form 
that the negative portion of the characteristic may be exactly divisible 
by the divisor, with — 10 as the quotient. 

Thus, to divide 8.5449 — 10 by 3, we write the logarithm in the 
form 28.5449 - 30 ; dividing this by 3, the quotient is 9.5150 - 10. 

85. Arithmetical Complement. 

The Arithmetical Complement of the logarithm of a num- 
ber, or, briefly, the Cologaritlim of the number, is the loga- 
rithm of the reciprocal of that number. 

Thus, colog 409 = log -^ = log 1 - log 409. 

log 1 = 10. - 10 (Note 2, § 84.) 
log 409= 2.6117 
.-. colog 409= 7.3883-10. 

Again, colog .067 = log — — = log 1 — log .067. 

logl= 10. -10 
log .067 = 8.8261 - 10 
.-. colog .067= 1.1739. 

It follows from the above that the cologarithm of a number 
may be found by subtracting its logarithm from 10 — 10. 

Note. The cologarithm may be obtained by subtracting the last 
significant figure of the logarithm from 10, and each of the others 
from 9, — 10 being written after the result in the case of a positive 
logarithm. 



Logarithms. 6i 

86. Example. Find the value of ^ F;^i7^* 

^ 8. 709 X. 0946 ^V 8.709 .0946/ -- 

= log . 51384 + log -i— + log 



8.709 *" .0946 
= log .51384 + colog 8.709 + colog .0946. 

log.51384 = 9.7109- 10 

colog 8.709 = 9.0601 -10 
colog .0946 = 1.0241 

9.7951 - 10 = log. 6239. 

It is evident from tlie above example that the logarithm 
of a fraction either of whose terms is the product of factors, 
may be found by the following rule : 

Add together the logarithms of the factors of the numerator, 
and the cologarithms of the factors of the denominator. 

EXAMPLES. 

Note. A negative number has no common logarithm (§ 62, Note). 

If such numbers occur in computation, they should be treated as if 
they were positive, and the sign of the result determined irrespective 
of the logarithmic work. 

Thus, in Ex. 3, § 87, the value of 439.2 x (- 7.1367) is obtained by 
finding the value of 439.2 x 7.1367, and putting a negative sign before 
the result. See also Ex. 33. 

87. Find by logarithms the values of the following : 

1. 3.145 X .6839. 4. (- 9.0654) x (- .010785). 

2. 847.6 X .02287. 5. .36552 x .025208. 

3. 439.2 X (- 7.1367). 6. - .0019036 x 57.143. 

7 4^6.7 Q - .2709 .. 8062.4 

76.52' ' .08683 * * 9.5073* 

g 1.0548 ^Q 6.802 ^^ .0001798 



34.96 .0051264 - .033166 



62 



Plane Trigonometry. 



13. 



38.961 X .695 
4994 X .0045' 



15. 



(- .87028) X 37 
(-.0659) X (-42.32)" 



j^ 715 X (- .02416) ^g .08214 x (- 73.4) 

(-.516) X 142.07' ■ .84 X 2808.7 



17. (7.795)^ 22. (.095129)1 

18. (.8328)^ 23. (.00010594)^ 

19. (-25.144)3. 24. V5. ' 

20. (.01)^. 25. </2. 

21. (-964.8)^. 26. V^^. 

2^5 



27. VIOO. 

28. ^"1995. 



29. V.072563. 



30. V.0026139. 

31. a/ -.00095174. 



32. Find the value of 



3* 



By § 86, 
log — -: 



log 2 + log \/5 + colog 3 6 



= log 2 + i log 5 + 1 colog 3. 

log2 = 0.3010 
log 5 = 0.6990 ; divide by 3 = 0.2330 

colog 3 = 9.5229 - 10 ; multiply by f = 9.6024 - 10 



0.1364 = log 1.369. 



33. Find the value of 



3 - .032< 
\ 7.962 



'-.03296 
.962 



log ■^- 



03296 ^ ^ J ^03296 ^ ^ .^ ^gg^g _ , ^^qq2\ 
7.962 ' '' 7.962 ' ^ ^ & ^ 



log .03296 = 8.5180 -10 
log 7.962 =0.9010 



3)27.6170 - 30 

9.2057 -10 = log. 1606. 



Kesult, - .1606. 



Logarithms. 
Find the values of the following : 
34. 4' X 7\ 



63 



39. (-Mooy. 

6928/ 



35. t 



\46 



40. ^ 
41. 



276.9 
940 * 

5^ 



44. -^Sx-^Sxa/T. 
45. 



76.1x.0593 \f 



1.307 



37. 



38. 



(.001)^ 

^7 ' 

v:o8 

(-10)*' 



42. 



-.1 

-J/iooo 



46.^" 
47 



75.44 



31.4 X .415 

-t/.0009657 



43 



(- -6? 

6/3 5/7 
• \5^V8- 



V.0049784 



4g^ -(.25693)-^ 
(-.8346)^ 



49. (25.467y«x(- .052)12. ^^ -</- .7664 x 1.2809 

50. \/5106.5 X .00003109. 

51. (837.5 X .0094325)1 

52. (4.8672)^ x (.17544)^ 
V3:929 X ^65M 



55. 



(.00259)^ 

^"^05287 



'.374 X V.0078359 



53. 



^721.33 



56. 



V.04142 X (- .947^) 



38.014 



57. .083184 X (.2682)^ x (56.1)2. 
58. 



.0005616 X a/424.65 



59. 



60. 



61. 



(6.73)* X (.03194)^ 
485.7 X ( .0 7301)^ x ^/MS 
(9.1273)« X (.7095)* 

(- .95048)^ X (8473)3 ' 
(-2080.9) x^:0572_ 

•^-.003012x1.955 
(- .843)« X -n/17959 x (- 560.6)* 



4 



64 Plane Trigonometry. 

EXAMPLES IN THE USE OF TRIGONOMETRIC 
TABLES. 

(For directions, see pages 4 to 8 of the Introduction to the 
author's New Four Place Logarithmic Tables.) 

88. Tables of Logarithmic Sines, Cosines, etc. 

Find the values of the following : 

1. log tan 35° 39'. 6. log sin 30° 37.2'. 

2. log sin 61° 58'. 7. log cos 55° 21' 48". 

3. log cot 12° 34'. 8. log cot 48° 3' 43". 

4. log cos 26° 56'. 9. log sec 80° 7'. 

5. log tan 82° 3'. 10. log esc 65° 12'. 

Find the angles corresponding in the following : 

11. log tan = 0.9164. 16. log cot = 0.2154. 

12. log cos = 9.9221 - 10. 17. log sin = 9.1891 - 10. 

13. log sin = 9.8619 - 10. 18. log tan = 8.9668 - 10. 

14. log cot = 9.4700 - 10. 19. log esc -= 0.1888. 

15. log cos = 9.2204 - 10. 20. log sec = 0.4032. 

Tables of Natural Sines, Cosines, etc. 

Find the values of the following : 

21. sin 17° 13'. 23. tan 35° 7'. 

22. cos 75° 38'. 24. cot 68° 46'. 

Find the angles corresponding in the following : 

25. sin = .7385. 27. tan = 1.1897. 

26. cos = .9280. 28. cot = 1.8207. 



Solution of Right Triangles. 



6S 



VI. SOLUTION OF RIGHT TRIANGLES. 

89. The elements of a triangle are its three sides and its 
three angles. 

We know by Geometry that a triangle is, in general, com- 
pletely determined when three of its elements are known, 
provided one of them is a side. 

The solution of a triangle is the process of computing the 
unknown from the given elements. 

90. To solve a rigJit triangle, two elements must be given 
in addition to the right angle, one of which must be a side. 

The various cases which can occur may all be solved by 
aid of the following formulae : 




sin^ = 
sin B = 



COS A = 



cos B 



tan^ 
tan 5 



Wheyi the given elements are a side and an 



91. Case I. 

angle. 

The formula for computing either of the remaining sides 
may be found by the following rule : 

Take that function of the angle which involves the given side 
and the required side. 

1. Given c = 23, B = 21° 33'. Find a and h. 
In this case, the formulae to be used are 
cos 5 



-, and sm B = -■ 



Whence, 



c cos B, and h 



c 
G sin B. 



(A) 



66 Plane Trigonometry. 



Solution by Natural Functions. 
a = 23 X cos 21° 33' = 23 X .9301 = 21.39. 
5 = 23 X sin 21° 33' = 23 x .3673 = 8.448. 

Solution by Logarithms. 

Taking the logarithms of both members, in formulae (A), 

log a = log c + log cos B, and log 6 = log c + log sin B. 

logc = 1.3617 logc = 1.3617 

log cos 5 = 9. 9685 - 10 log sin B = 9. 5651 - 10 

log a = 1.3302 log 6 = 0.9268 

a = 21.39. 6 = 8.448. 

Note. In examples under Case I. in which the given sides are 
numbers of not more than two significant ^figures, and the operations 
indicated involve only multiplication, it is usually shorter to employ 
Natural Functions. 

In such a case, the results cannot be depended upon to more than 
four significant figures. 

2. Given a = .2369, A = 67° 18'. Find b and c. 

In this case, tan A = -, and sin A = -. 

h G 

Whence, ^ ^ 



tan A sin A 

By logarithms, 

log h = \oga — log tan A^ and log c = log a — log sin A. 

log a = 9.3727 - 10 log a = 9. 3727 - 10 

log tan ^ = 0.3785 log sin A = 9. 9650 - 10 

log h = 8.9942 - 10 log c = 9.4077 - 10 

b= .09868. c= .2557. 

92. Case II. Wlieji both given elements are sides. 

First calculate one of the angles by aid of either formula 
involving the given elements, and then compute the remain- 
ing side by the rule of Case I. 



Solution of Right Triangles. 67 

Ex. Given h = .1512, c = .3081. Find A and a. 

We first find A by the formula cos J. = -, and then a by the 

c 
formula sin ^ = -, or a = c sin ^. 

c 

By logarithms, 

log cos J. = log 5 — log c, and log a = log c + log sin A. 

log h = 9.1796 - 10 log c = 9.4887 - 10 

logc= 9.4887 -10 logsin^ = 9.9401 -10 

log cos A = 9.6909 - 10 log a - 9.4288 - 10 

^ = 60° 36. 4'. a= .2684. 

93. In the Trigonometric solution of an example under 
Case II., it is necessary to first find one of the angles, and 
the remaining side may then be calculated. 

But it is possible to compute the third side directly, with- 
out first finding the angle, by Geometry. 

Thus, in the example of § 92, we have 
a2 + 52 ^ ^2_ 



Whence, a = Vc' - 6^ = V(c + 6) (c - h). 
By logarithms, log a = ^ [log (c + 6) + log (c — 6)]. 

c + 6 = .4593 ; log = 9.6621 - 10 
c - 5 = .1569 ; log = 9.1956 - 10 
2 ) 18.8577 -"20 
loga = 9.4289 -10 
a= .2685. 

If the given sides are a and h, the expression for c is 
Va^ + &^ which is not adapted to logarithmic computation. 
In such a case, it is usually shorter to proceed as in § 92. 

EXAMPLES. 
94. Solve the following right triangles : 
H. Given A = 15°, c = 7. 3. Given B = 50°, b = 20. 
2. Given B = 68°, a = 5. 4. Given a = .35, c = .62. 



68 



Plane Trigonometry. 



5. Given a = 21, 5 = 42. 8. Given 6 = 586, c= 763. 

6. Given A = 38°, a = 8.09. 9. Given A = 9°, b= 937. 

7. Given B = 65% c = .014. 10. Given a = 3.41, b = 2.87. 

11. Given ^ =3 31° 50', a = 48.04. 

12. Given ^ = 46° 15', c = 5280. 

13. Given 6 = .0469, c = .0515. 

14. Given 5 = 79° 28', 6 = 842. 

15. Given B = 67° 47', c = .00954. 

16. Given A = 43° 30', b = 26185. 

17. Given a = 3402, 6 = 2317. 

18. Given 5 = 82° 6', a = .08937. - 

19. Given 6 = 578.9, c = 2492. 

20. Given A = 26° 12', c = .4694. 

21. Given B = 14° 53', 6 = 1353. 

22. Given 5 = 43° 24', a = .89658. 

23. Given a = 99.46, c = 156.8. 

24. Given A = 62° 44', 6 = 4.2492. 

25. Given A = 74° 17', a = .000020386. 

26. Given B = 29° 56', c = .00078144. 

27. Given a = 63827, c = 92275. 

28. Given A = 58° 39', c = 35.733. 

29. Given 5 = 35° 8', 6 = 17269. 

30. Given ct = .0067239, 6 = .0038453. 

Solve the following isosceles triangles, in which A and B 
are the equal angles, and a, 6, and c the sides opposite 
angles A, B, and C, respectively : 

31. Given A = 71°, 6 = 39. 

32. Given B = 36° 40', c = .4688. 




Solution of Right Triangles. 69 

33. Given (7= 83° 52', & = 710.6. 

34. Given a = 6875, c = 11318. 

35. Given B = 29° 7', a = 2.569. 

36. Given A = 54° 39', c = 1.7255. 

37. Given C = 135° 26', c = .06377. 

MISCELLANEOUS PROBLEMS. 

95. If ^C is the diagonal of rectangle ABCD, and the 
side AB is horizontal and BC vertical, 
Z BAC is called the cmgle of elevation 
of point C from point A, and Z ^(7i> 
the angle of depression of point A from 
point (7. 

96. 1. From the top of a lighthouse, 150 feet above the 
sea, the angles of depression of two boats, in line with the 
lighthouse, are observed to be 12° and 30°, respectively. 
Find the distance between the boats. 

Let A be the position of the first boat, A' of the second, B the top 
of the lighthouse, and C its foot. 

Then, 
AA> = AC- A'C ^ BC cot A- BC cot BA'C 
= 150 (cot 12° -cot 30°) 
= 150(4.7046-1.7321) 
= 150 X 2.9725 = 445.9 ft. 

2. From the top of a lighthouse, 250 feet above the sea, 
the angle of depression of a buoy is observed to be 31°. 
Find the horizontai distance of the buoy. 

3. If the radius of a circle is 834, what is the length of a 
chord which subtends an arc of 46° ? 

4. A regular hexagon is circumscribed about a circle 
whose diameter is 59. Find the length of its side. 




yo Plane Trigonometry. 

5. Find the angle of elevation of a road wMcli rises a 
distance of 349 feet in a horizontal distance of five-eighths 
of a mile. 

6. A regular polygon of nine sides is inscribed in a circle 
whose diameter is 6S. Find the length of its side. 

7. How far from the top of a tower 121 feet in height 
mnst an observer stand so that the angle of elevation of its 
top may be 23° ? 

8. A regular polygon, whose side is 7.6 and angle 144°, is 
circumscribed about a circle. Find its radius. 

9. Find the angle of elevation of the sun when a monu- 
ment whose height is 214.8 feet casts a shadow 167.4 feet 
in length. 

10. Find the length of the diagonal of a regular pentagon 
whose side is 9.437. 

11. At a distance of 41.6 feet from the base of a tower, 
the angle of elevation of its top is observed to be 59° 36'. 
Find its height. 

12. The middle point of a chord of a circle, 24 units in 
length, is distant 7 units from the middle point of its sub- 
tended arc. How many degrees and minutes are there in 
the arc ? 

13. If the diameter of a circle is 6374, find the angle at 
the centre subtended by an arc whose chord is 2138. 

14. If the radius of a circle is 9.54, and the distance from 
the middle point of a chord to the middle point of its sub- 
tended arc is 3.87, how many degrees and minutes are there 
in the arc ? 

15. From the top of a tower, the angle of depression of 
the extremity of a horizontal base line, 236.1 feet in length 
measured from the foot of the tower, is observed to be 
29° 48'. Find the height of the tower. 



Solution of Right Triangles. 71 

16. A chord of a circle, whose length is 14.95, subtends 
an arc of 135° 52'. What is the distance from the middle 
point of the chord to the middle point of the arc ? 

17. If a vertical pole casts a shadow which is three-fourths 
its own length, what is the angle of elevation of the sun ? 

18. The radius of the inscribed circle of an equilateral 
triangle is .307. Eind its perimeter, and the diameter of 
the circumscribed circle. 

19. The side of a regular octagon is 23.68. Find the 
radii of its inscribed and circumscribed circles. 

20. A chord of a circle subtends an arc of 70° 24'. If the 
length of the chord is 853.4, find the radius of the circle. 

21. At a point 250 feet from the foot of a cliff surmounted 
by a lighthouse, the angle of elevation of the top of the 
lighthouse is 50°, and of its foot 30°. Find the height of 
the cliff, and of the lighthouse. 

22. From the top of a cliff 378 feet above the sea, the 
angles of depression of two boats, in line with the observer, 
are observed to be 11° 50' and 29° 20', respectively. Find 
the distance between the boats. 

23. At a distance of 169 feet from the foot of a tower 
surmounted by a pole, the tower subtends an angle of 35°, 
and the pole an angle of 12°. Find the length of the pole. 

24. How many degrees and minutes are there in the arc 
included between two parallel chords, on the same side of 
the centre of a circle, whose distances from the centre are 5 
and 7, respectively, the radius of the circle being 11 ? 

25. From the top of a tower, the angle of depression of a 
stake is 31° 29'. What will be the angle of depression of 
the stake from a point half way to the top ? 

26. The diagonal of a regular pentagon is 43.92. Find 
the radius of its inscribed circle. 



72 Plane Trigonometry, 

27. A railway runs from AtoB,Si horizontal distance of 
1250 feet, at an angle of elevation of 8° 12', and then from 
jB to C, a horizontal distance of 375 feet, at an angle of elevar 
tion of 7° 26'. How many feet is C above the plane oi A? 

28. From a point 200 feet from the foot of a tower sur- 
mounted by a pole, the angle of elevation of the top of the 
pole is 38°; from a point 150 feet further, the angle of 
elevation of the foot of the pole is 22°. Find the height 
of the pole. 

29. If the radius of the earth is 3956 miles, find the 
radius in miles of the arctic circle, latitude 66° 32' IST. 

30. If the diameter of the earth is 7912 miles, what is 
the distance of the remotest point of the surface visible 
from the top of a mountain, l-i- miles above the sea ? 

31. A flagpole 23 feet long surmounts a tower whose 
height is 98 feet. What angle does the flagpole subtend at 
a point on the ground 315 feet from the base of the tower ? 

32. An observer notes that a spire bears due north from 
him, the angle of elevation of its top being 22° 17'. On 
going due east 550 feet, the spire bears 49° west of north. 
What is the height of the spire ? 

33. A regular pyramid stands on a square base, whose 
side is 50 feet. Each side of the base makes an angle of 
69° with the lateral edge. Find the altitude of the pyramid. 

34. A vessel is sailing due north at a uniform rate of 
speed. At 7.30 a.m., a lighthouse is observed to bear 70° 
west of north ; at 8 a.m. it is due west, at a distance of 12 
miles. Find the distance and bearing of the lighthouse 
at 9.30 A.M. 

FORMULA FOR THE AREA OF A RIGHT TRIANGLE. 

97. Case I. Given the liyiootenuse and an acute angle. 



Solution of Right TriangleSo 73 

B 




b 

Denoting tlie area by K, we have by Geometry, 

2K=zab. 
But by § 5, a = c sin A, and b = c cos A. 
Whence, 2 K=c^ sin Aco^A = \c^ sin 2 A, by (22). 
Then, 4 /iT = c^ sin 2 ^. (32) 

In like manner, 4 /r= c^ sin 2 jB. (33) 

Case II. Given an angle and its opposite side. 

By § 2, b = aGotA. 

Whence, 2 K=a x a cot A = a^ cot J.. (34) 

In like manner, 2 K=b^ cot B. (35) 

Case III. Given an angle and its adjacent side. 

By § 2, b = a tan B. 

Whence, 2 K=a x a tan 5 = a^ tan 5. (36) 

In like manner, 2 K=b^ tan A. (37) 

Case IV. (Twe?i the hypotenuse and another side. 
Since a^ + 5^ = c^, we have 

2 K= ab = a^(? — a^ — a^(c + a) (c — a). (38) 



In like manner, 2K= b-y/ic -\-b){G — b). (39) 

Case V. Given the two sides about the right angle. 

In this case, 2 K= ab. (40) 



74 Plane Trigonometry. 

EXAMPLES. 

98. 1. Given c = 10.36, B = T5°; find the area. 

By (33), 4ir=c2sin2^. 

Whence, log (4 K) = 2 log c + log sin 2 B. 

log c = 1.0153 ; multiply by 2 = 2.0306 

2^=150°; log sin = 9.6990 -10 

log (4^) =1.7296 

.-. 4ir= 53.65, and ^=13.41. 

Note. To find log sin 150°, take either log cos 60° or log sin 30°. 
(See page 7 of the Introduction to the author's New Four Place 
Logarithmic Tables.) 

Find the areas of the following right triangles : 

2. Given A = 46°, a = 2.717.^ 

3. Given B = 35° 16', a = .557. 

4. Given a = 283.17, b = 94.93. 

5. Given b = 4.564, c = 7.176. 

6. Given A = 53° 9', c = 13.84. 

7. Given A = 20° 57', b = .05027. 

8. Given a = .0861, c = .4806. 

9. Given B = 67° 48', c = 67.409. 

10. Given B = 75° 34', b = .0032056. 

11. Given ^ = 81° 23', c = 195.84. 



General Properties of Triangles. 



IS 



VII. GENERAL PROPERTIES OF 
TRIANGLES. 

99. In any triangle, the sides are proportional to the sines 
of their opposite angles. 

I. To prove a : 6 = sin ^ : sin B. (41) 





There will be two cases, according as angles A and B are 
both, acute (Fig. 1), or one of them obtuse (Fig. 2). 
In each case, draw line CD perpendicular to AB. 



Then in each figure, 
Also in Fig. 1, 
And in Fig. 2, 



CD = b sin A (§ 5). 
CD = a sin B. 
CD = a sin CBD 



= a sin (180° -B) = a sin B 

(§ 32). 
Then in either case, b sin ^ = a sin B. 



Whence by the theory of proportion, 

a : 6 = sin ^ : sin B. 

In like manner, b : c = sin B : sin C, 

and c : a = sin (7 : sin A. 



(42) 
(43) 



100. In any triangle, the sum of any tico sides is to their 
difference as the tangent of half the sum of the opposite angles 
is to the tangent of half their difference. 

By (41). a : 6 = sin J. : sin B. 



76 Plane Trigonometry. 

Whence by composition and division, 

a-{-b : a — b = sin^H-sin_B : sin^— sinjB. 
a -\-b sin A + sin B 



Or, 



a ~ b sin A — sin B 



^ sin ^ + sin ^ _ tani(^ + ^) , (21). 

sin^-sinjB tan-i-(^-5)' ^ ^ ^ 



Whence, 



a -]- b ^ t2,i\\ {A -\- B) 
a — b tan ^{A — B) 



T n b^C' tani(5+C) 
In like manner, — ' — = 2_a ! z 

' b-c tan i (J3 - G) 



and 



c4- a _ tani((7 + J.) 
c — a tan-i-((7 — ^) 



(44) 
(45) 
(46) 



101. In any triangle, the square of any side is equal to the 
sum of the squares of the other two sides, minus twice their 
product into the cosine of their included angle. 



I. To x^rove a^ = W -\- c^ — 2bc cos A. 

Case I. When the included angle A is acute. 
C 



(47) 





There will be two cases, according as angle B is acute 
(Fig. 1), or obtuse (Fig. 2). 

In each case, draw line CD perpendicular to AB. 

In Fig. 1, BD = c- AD, and in Fig. 2, BD = AD - c. 

Squaring, we have in either case, 



BD^ = AD^ + c'--2cx AD. 



General Properties of Triangles. 

Adding CD to both members, 

Blf -hCI? = ad' +CD'-\-c'-2cx ad. 
But, Bff +Clf = o?, and Aff -{- Ud' = h\ 
Also, by § 5, AD = h cos A. 

Whence, o? = h^ -{- c^ — 2bc cos A. 

Case II. When the included angle A is obtuse. 

Q 



77 




Draw line CD perpendicular to AB. 
We have BD = AD -\- c. 

Squaring, and adding CD to both members, 

BD'-hCD' = ad' +CD^ + c^ + 2cxAD. 
But, BD'-{-CD' = a', and AD' -{- CD' = h\ 
And by § 5, 
AD = h cos CAD = h cos (180° - ^) = - 6 cos ^ (§ 32). 

Whence, a^ = b^ -\- c^ — 2 be cos A. 

In like manner, b^ = c^ -\- a^ — 2ca cos B, (48) 

and c^ = a^ -\-b^ — 2 ab cos C (49) 

102. To ex^jress the cosines of the angles of a triangle in 
terms of the sides of the triangle. 

By (47), a^==W + c'-2bc cos A. 

Transposing, 2 be cos A = V^ -\- c^ — a^. 



78 Plane Trigonometry. 

Whence, cos A = ^' + ^' - ^ . /gg) 

2 be ^ 

In like manner, cos B = — ^^^I , (51) 

2ca ^ 

and cos C = ^' +J>' - ^l (52) 

2ab 



103. jPo express the sines, cosines, and tangents of the half 
angles of a triangle in terms of the sides of the triangle. 

By (50), 

2 5c 2 he 

Whence by (28), 

o • 1 A a^ — (b — cY 

2 sin^ ^A= ^ ^• 

^ 2bc 

Or, sinH^- ^^~^ + '^^^' + ^~'^ ' 

^ Abe 

Denoting the sum of the sides, a + 6 + c, by 2 s, we have 
a - b -\- c = (a -{- b + c) - 2b == 2 s - 2b = 2(s - b), 
and a + 6 — c = (a + 5 + c) — 2c = 2s — 2c = 2(s — c). 
Whence, sin^ i ^ = 4(.-6)(.-c) , 



Or, 



sinl^=.J5^ME5. (53) 



In like manner, sin i B =\h^ — ^^-^ ^> (54) 

^ ca 



smiC=^J ^'-''^^'-^^ ' (55) 



and _ 

ab 



General Properties of Triangles. 79 

Again, by (50), 

1 + cos ^ = 1 H — = — 

2 be 2 be 

Whence by (29), 

2cosH^ 



(P ^ cf - 0} 



9 



be 



Or, cos^^- ^^^ 

But, 6 + c + a = 2s, 
and b -\- e — a = (b -\- G + a) — 2 a = 2 {s — a). 



Whence, 



30S- 1 A ■■ 


_46 

/■ 


4. be 


± 


COS i J. : 


s(s — 
be 


a). 



Or, cos i A = A/H^^^. (56) 



In like manner, cos -J--B =^— h (57) 



ea 



and cos i (7 = ^^ii— ^. (58) 

Dividing (53) by (56), we have. 



sin^A ^ l (s-b)(s-e) I be 
cos i J. ^ be ^s{s — a) 



Whence by (4), tan i ^ = J (^ ^)(s-e) ^ ^^^^ 

s is — Ct) 



In like manner, tani^^^ r ^^'^^^'^~^\ (60) 

2 \ s(s_5) ^ ^ 



and UniC=J ('-'^^('-^\ (61) 



8o 



Plane Trigonometry. 



Note. Since each angle of a triangle is less than 180°, its half is 
less than 90° ; hence, the positive sign must be taken before the radi- 
cal in each formula of § 103. 

FORMULA FOR THE AREA OF AN OBLIQUE TRIANGLE. 

104. Case I. Given two sides and their included angle. 
I. When the given parts are h, c, and A. 

A ^ 





There will be two cases, according as A is acute (Fig. 1), 
or obtuse (Fig. 2). 

In each case, draw line CD perpendicular to AB. 
Then denoting the area by K, we have by Geometry, 

2K=cxCD. 
But in Fig. 1, CD = b sin A (§ 5). 

And in Fig. 2, CD = b sin CAD 

= h sin (180° - ^) = 6 sin ^ (§ 32). 



Then in either case, 

2 K= hc^xuA. 

In like manner, 2 K= ca sin B, 

and 2 K= ah sin C. 

Case II. Given a side and all the angles. 
I. When the given parts are a. A, B, and G. 

By (64), 2K=absmC 



(62) 
(63) 
(64) 



General Properties of Triangles. 8i 

-D i. u /y,-,\ b sin B J a sin B 

But by (41), - = — , or b=—. — • 

^ ^ ^' a smA' sin A 

Whence, 2K=ax ^L$^ x sin G 

smA 



a^ sin B sin O 
sin^ 



(65) 



T Ti o T^ b^ sin C sin A /--^ 

In like manner, 2K= -. ? (66) 

' sm^ ^ ^ 

and ^j^^l^nA^^ ^ 

sm C 



Case III. Given the three 

By (62), 2 ^= 6c sin ^ = 2 6c sin i ^ cos i A, by (22). 

Dividing by 2, and substituting the values of sin i A and 
cos I" -4 from (53) and (56), we have. 



\ he, \ he 



= Vs(s-a)(s-b)(s-c). (68) 



82 Plane Trigonometry. 



VIII. SOLUTION OF OBLIQUE TRIANGLES 

In the solution of oblique triangles^ we may distinguish 
four cases. 

105. Case I. Given a side and any two angles. 

The third angle may be found by Geometry, and then by 
aid of § 99 the remaining sides may be calculated. 

The triangle is possible for any values of the given ele- 
ments, provided the sum of the given angles is < 180°. 

1. Given & = 20, A = 104°, ^ = 19° ; find C, a, and c» 
We have C = 180° -(A-{-B)= 180° - 123° = 57°. 



By § 99, 


a _ sin ^ 
h sin B 


;, and ^ 


_ sin C 
sin^ 




Then, 


a = & sin 


A CSC B, 


and c = 6 sin esc B. 




Whence, 


log a = log 6 


+ log sin 


A + log CSC B, 




d 


log c = log 6 


+ log sin 


C + log CSC B. 




log 6: 


= 1.3010 




log & = 1.3010 




log sin A = 


= 9.9869 - 10 




log sin a =9.9236 - 


-10 


log CSC B - 


= 0.4874 




log CSC ^ = 0.4874 




loga = 


:z 1.7753 


logc = 1.7120 




a - 


z 59.61. 




c = 51.52. 





Note. To find the log cosecant of an angle, subtract the log sine 
from 10 — 10. To find log sin 104°, take either log cos 14° or log sin 
76°. (See page 7 of the author's New Four Place Logarithmic 
Tables.) 

EXAMPLES. 

Solve the following triangles : 

2. Given a = 180, A = 38°, B = 75° 20'. 

3. Given b = 8.19, B = 52°, C = 109°. 

4. Given c = .0246, ^ = 83° 30', 5 = 38° 50'. 



Solution of Oblique Triangles. 83 

5. Given 6 = 67.13, A = 26° 18', C = 44° 35'. 

6. Given c = .45924, ^=74° 43', C=61°29'. 

7. Given a = 3024, B = 133° 34', C = 22° 57'. 
(For additional examples under Case I, see § 112.) 

106. Case II. Given two sides and their included angle. 

Since one angle is known, the sum of the remaining angles 
may be found, and then their difference may be calculated 
by aid of § 100. 

Knowing the sum and difference of the angles, the angles 
themselves may be found, and then the remaining side may 
be computed as in Case I. 

The triangle is possible for any values of the data. 

1. Given a = S2, c = 167, 5 = 98°; find A, C, and b. 
By Geometry, + ^ = 1 80° -5 = 82°. 



By § 100, 



c + a _ tan-i (C+ ^) 
c — a tan ^ (^C — A) 



Or, tani((7-^):=^^^tani((7 + ^). 

Then, 
log tan ^ ( C - J.) = log (c — a) + colog (c + a) + log tan 1 ( + A). 

c-a = Sb. log =1.9294 

c + a = 249. colog = 7.6038 - 10 

1(0 + ^)= 41°. log tan = 9.9392 - 10 

log tan ^(C-A)= 9.4724 - 10 
i(C-^)=16°31.7'. 
Then, (7 = K<^'+ -4)+ K^- ^)= 67°31.7', 

and A = l(C + A)- ^(C- A) = 24:° 28.SL 

To find the remaining side, we have by § 99, 

I a sin B ■ n a 

b = —, = a sm J5 esc A 

sinA 



84 Plane Trigonometry. 

Whence, log 6 = log a + log sin B + log esc A. 

loga = 1.9138 
logsin5 = 9.9958 -10 
log CSC ^ = 0.3828 

log & = 2.2924 
6 = 196.1. 

EXAMPLES. 

Solve the following triangles : 

2. Given a = 67, c = 33, B = 36°. 

3. Given a = 986, 6 = 544, 0=134°. 

4. Given b = .149, c = .427, A = 71°. 

5. Given a = 3.95, b = 6.64, = 68° 30'. 

6. Given a = 2937, c = 6185, B = 55° 46'. 

7. Given 6 = .01292, c = .00286, ^ = 26° 32'. 
(For additional examples under Case II., see § 112.) 

107. Case III. Given the three sides. 

The angles might be calculated by the formulae of § 102 ; 
but as these are not adapted to logarithmic computation, it 
is usually more convenient to use the formulae of § 103. 

Each angle should be computed trigonometrically ; for we 
then have a check on the work, since their sum should be 
180°. 

If all the angles are to be computed, the tangent formulae 
are the most convenient, since only four different numbers 
occur in the second members. 

If but one angle is required, the cosine formula involves 
the least work. 

The triangle is possible for any values of the data, pro- 
vided no side is greater than the sum of the other two. 



Solution of Oblique Triangles. ' 85 

If all the angles are required, and the tangent formulae 
are used, it is convenient to modify them as follows ; by (59), 



tan 1^ ^ V(--^') ^r't"'^ = J->-«) (^-^) (--<-). 

' S (S — Clj S — Ct ' s 



Denoting ^^^ ~ ^^) ^^ " ^) (^ " ^) by r, we have 



tani J[ 



— a 



In like manner, tan \B = , and tan 1 G 



s — c 



1. Given a = 2.5, b = 2.S, = 2.2-, find A, B, and O. 
Here, 2 s = a -\- b -\- c = 7.5, and s = 3.75. 
Then, s — a = 1.25, s — b = .95, s — c = 1.55. 
By logarithms, 

log r = 1 [log (s - a) + log (s - 5) + log (s - c) + colog s]. 
Also, log tan 1 ^ = log r — log (s — a), 

log tan ^B = logr — log (s — 6) , 

log tan 1 C = log r — log (s — c). 

log (s - a) = 0.0969 log r = 9.8455 - 10 

log (s - &) = 9.9777 - 10 log (s-b)= 9.9777 - 10 

log (s - c) = 0. 1903 log tan 1 5 = 9.8678 - 10 

colog 5 = 9.4260- 10 



J5 = 36°24.6'. 
5=72° 49.2'. 



2 )19.6909 - 20 
log r = 9.8455 - 10 
log (s- a) = 0.0969 log r = 9.8455 - 10 



tani^ = 9.7486- 10 



(s-c) = 0.1903 



J^ = 29°16.3'. 

^ = 58° 32.6'. JC'=24°19.7'o 

Check, A + B+ C= 180° 1.2'. 



tan ^0=9.6552- 10 

C'=24°19.7'„ 

= 48° 39.4'. 



86 Plane Trigonometry. 

2. Given a = 7, b = ll, c = 9.6 ; find B. 
By §103, cos^B = ^]'^'~^'^. 

CO, 

Or, log cos 1 5 = i[log s + log (s - 5) + colog c + colog a]. 
Here, 2s = 27.6 ; whence, s = 13.8, s - & = 2.8. 

logs = 1.1399 

log (s- 6) = 0.4472 

colog c = 9.0177 -10 

colog a = 9.1549 -10 

2 )19.7597 -20 

log cos 1^ = 9.8799 - 10 

|5 = 40°40.9', and 5 = 81" 21.8'. 

EXAMPLES. 

Solve tlie following triangles : 

3. Given a = 5, h = l, c=zQ>. 

4. Given a == 10, 6 = 9, c = 8. 

5. Given a = M, b = A3, c = .89. ■ 

6. Given a = 70.5, b = 56.2, c = 63.9 ; find A. 

7. Given a = .0292, & = .0185, c = .0357; find 5. 

8. Given a = 302, b = 427, c = 674 ; find C. 
(For additional examples under Case III., see § 112.) 

108. Case IV. Given two sides, and the angle opposite to 
one of them. 

It was stated in § 89 that a triangle is in general com- 
pletely determined when three of its elements are known, 
provided one of them is a side. The only exceptions occur 
in Case IV. 



Solution of Oblique Triangles. 



87 



To illustrate, let us consider the following example : 
Given a = 52.1, & = 61.2, A = 31" 26'-, find B, C, and c. 

By §99, sin^^6^ ^^ sin 5 = ^il^. 

sin A a a 

Whence, log sin B = \ogb + colog a + log sin A. 

log6 = 1.7868 

cologa = 8.2832 - 10 

logsin^ = 9.7173 -10 

log sin^ = 9.7873- 10 

B = 37° 47.5', from the table. 

But in finding tlie angle corresponding, attention must be paid to 
the fact that an angle and its supplement have the same sine (§ 32). 

Therefore another value of B will be 180° - 37° 47.5', or 142° 12.5' ; 
and calling these values ^i and B2, we have 

Bi = 37° 47.5', and B2 = 142° 12.5'. 

The reason for the ambiguity is at once apparent when we attempt 
to construct the triangle from the data. 




Bi D 



We first lay off angle DAF=Sr26', and on AF take AC =61.2. 
With O as a centre, and a radius equal to 52.1, describe an arc cutting 
AD at Bi and B2. Then either of the triangles ABiC or AB^C 
satisfies the given conditions. 

The two values of B which were obtained are the values of angles 
AB\C and AB^G^ respectively; and it is evident geometrically that 
these angles are supplementary. 

To complete the solution, denote angles ACBi and ACBq, by C\ and 
C2, and sides ABi and AB2 by C\ and C2, respectively. 

Then, Ci = 180° - (^ + ^1) = 180° - 69° 13.5' = 110° 46.5', 

and (72 = 180° - (J. + B2) = 180° - 173° 38.5' = 6° 21.5'. 



88 Plane Trigonometry. 

Again, by § 99, ^ = ^-H^, and ^ = ^J^^. 
a sin^ a sin A 

Whence, Ci = a sin Ci esc A^ and C2 = a sin C2 esc A 

loga = 1.7168 loga = 1.7168 

log sin Ci = 9.9708 - 10 log sin C2 = 9.0443 - 10 

log esc JL = 0.2827 log esc ^ = 0.2827 

log ci = 1.9703 log C2 = 1.0438 

ci = 93.40. C2 = 11.06. 

109. Whenever an angle of an oblique triangle is deter- 
mined from its sine, both the acute and obtuse values must 
be retained, unless one or both can be shown to be inadmis- 
sible ; hence there may sometimes be two solutions, some- 
times one, and sometimes none, in an example under Case IV. 

1. Let the data be a, h, and A, and suppose h < a. 

By Geometry, B must be < ^; hence, only the acvte value 
of B can be taken; in this case there is but one solution. 

2. Let the data be a, b, and A, and suppose b > a. 
Since B must be > A, the triangle is impossible unless A 

is acute. 

Again, since = -, and 6 is > a, sinB is > sin A 

sm^ a 

Hence, both the acute and obtuse values of B are > A, 
and there are two solutions, except in the following cases : 

If log sin B = 0, then smB=l (§ 68), and B = 90°, and 
the triangle is a rigJit triangle ; if log sin B is positive, then 
sin jB is > 1, and the triangle is impossible. 

The above results may be stated as follows : 

If, of the given sides, that adjacent to the given angle is 
the less, there is but one solution, which corresponds to the 
acute value of the opposite angle. 

If the side adjacent to the given angle is the greater, there 
are two solutions, unless the log sine of the opposite angle is 
or positive ; in which cases there are one solution (a right 
triangle), and no solution, respectively. 



Solution of Oblique Triangles. 89 

110. We will illustrate these points by examples : 

1. Given a = 7.42, h = 3.39, A = 105° 13' ; find B. 

Since 6 is < a, there is but one solution, corresponding to the acute 
value of B. 

By § 99, sin B = P^ILA. 

a 

log&= 0.5302 
colog«=: 9.1296- 10 
logsin^=: 9.9845- 10 



log sin 5 = 9.6443 — 10 
5 = 26° 9.6'. 

2. Given b = S, c = 2, C= 100° ; find B. 

Since & is > c, and C is obtuse, the triangle is impossible. 

3. Given a = 22.764, c = 50, A = 27° 4.8' 5 find C. 

We have, sin C = ^^HL^. 

a 

logc = 1.6990 
colog« = 8.6428 -10 
log sin^ = 9.6582 -10 
logsin (7= 0.0000 
Therefore, sin (7 = 1, and C= 90°. 

Here there is but one solution ; a right triangle. 

4. Given a =.83, 6 = .715, i? = 6r47'; find A 

We have, sin^ = ^^^^^ . 

b 

log a = 9.9191 -10 
colog & = 0. 1457 
log sin ^ = 9.9451 - 10 
log sin vl = 0.0099 
Since logsin J. is positive, the triangle is impossible. 



90 Plane Trigonometry. 

EXAMPLES. 
111. Solve the following triangles : 



1. 


Given a = 7.3, 


b = 6.6, 


^ = 56°. 


2. 


Given b = 86, 


c = 159, 


0=115°. 


3. 


Given & = 60.93, 


c = 76.09, 


5 = 133° 41. 


4. 


Given b = 3S, 


c = 48. 


B = 34°. 


5. 


Given a = .279, 


c - .227, 


(7 =65° 45'. 


6 


Given a = 3215, 


c = 6754, 


^ = 28° 26'. 


7. 


Given a = .06358, 


c = .08604, 


0=19° 14'. 


8. 


Given a = 186.7, 


b = 394.2, 


J5 = 114° 28'. 


9. 


Given a = .462, 


c = .647, 


A = 31° 7'. 



(For additional examples under Case IV., see § 112.) 

MISCELLANEOUS EXAMPLES. 
112. Solve the following triangles : 



1. 


Given 


a = 934, 


b = 756, 


0=73° 16'. ^ 


2. 


Given 


c = 8.706, 


5 = 38° 45', 


0=31° 59'. 


3. 


Given 


a = 61, 


6 = 85, 


c = 48. 


4. 


Given 


a = .425, 


c = .454, 


0=37° 9'. 


5. 


Given 


b = .0479, 


c = .0144, 


^ = 121° 28'. 


6. 


Given 


a = 7824, 


c = 3202, 


A = 140° 53'. 


7. 


Given 


b = .0005639, 


^ = 44° 24', 


5 = 116° 9'. 


8. 


Given 


a = 1.5, 


b = 1.3, 


c = 1.9. 


9. 


Given 


a = 576, 


b = 813, 


A = 23° 25'. 


10. 


Given 


b = 2615, 


c = 6086, 


A = 115° 10'. 


11. 


Given 


b = 9.874, 


c = 7.486, 


5 = 81° 47'. 


12. 


Given 


a = 71387, 


B = 42° 56', 


0=76° 7'. 



Solution of Oblique Triangles. 91 

13. Given b = 51.434, c = 47.955. C = 72° 54'. 

14. Given a = .008727, c = .007065, B = 84° 56'. 

15. Given a = .031, 6 = .024, c = .028. 

16. Given a = .19597, 6 = .13927, ^ = 45° 17'. 

17. Given a = 3.5374, 5 = 9.6036, A = 97° 46'. 

18. Given a = .40932, ^ = 53° 13', C=67°32'. 

19. Given a = 31.06, b = 51.49, C = 47° 43'. 

20. Given a = .019186, b = .033728, ^ = 125° 33'. 

21. Given a = 353.85, c = 579.42, 5 = 19° 37'. 

22. Given 6 = 24883, c = 20609, C = 48°6'. 

AREA OF AN OBLIQUE TRIANGLE. 
113. 1. Given a = 18.063, A = 96° 30', B = Sd°; find Kl 

By § 104, 2 K= «" sm -B sm C ^ ^2 gin ^ sin esc ^. 

sill J. 
Whence, 

log (2 K) = 2 log a + log sin 5 + log sin C + log esc A. 

Here, C = 180° - (^ + 5) = 48= 30'. 

log a = 1.2568 ; multiply by 2 = 2.5136 

logsin5 = 9.7586 - 10 
log sin C= 9.8745 -10 
logescJ. = 0.0028 



log (2 IT) = 2.1495 

2/f= 141.1, and ir= 70.55. 

EXAMPLES. 

Find the areas of the following triangles : 

2. Given a = 26 A, c = 47.9, B = 67°. 

3. Given a = 8.05, B = 65° 30', C = 81° 40'. 

4. Given a = 7, 6 = 9, c = 6. 



9^ Plane Trigonometry. 

5. Given c = .518, ^==67° 45', B = 37°19'. 

6. Given b = 15.32, c = 36.78, A = 105° 43'. 

7. Given 5-210.6, ^ = 32° 21', O = 108° 56'. 

8. Given c = .004096, A = 17° 4.5', = 46° 8'. 

9. Given a = .73, b = .55, c = .63. 

10. Given a = .0006854, b = .0009743, C = 61° 44'. 

11. Given a -3 7.219, ^ = 23° 33', JB = 124° 12'. 

12. Given a = 5.321, c = 8.467, 5 = 152° 51'. 

13. Given a = 39.5, b = 47.3, c = 50.8. 

14. Given b = 250.8, A = 77° 53', O = 56° 29'. 

15. Given & = . 19146, c = .42829, ^ = 59° 7'. 

16. Given a = .078, b = .091, c = .084. 

17. Given & = 109.41, ^ = 77° 46', 5 = 43° 32'. 

18. Given a = 5.7434, b = 8.6326, C = 129° 17'. 

19. Given a = 307.4, b = 351.9, c = 335.7. 

20. Given a = .0083214, A = S^°U', = 105° 23'. 

21. Given a = .064325, c = .033777, 5 = 141° 38'. 

MISCELLANEOUS PROBLEMS. 

114. 1. To find the distance of an inaccessible object A 
from a position B, I measure a base-line BG 675 feet in 
length, and observe the angles ABO and ACB to be 101° 17' 
and 36° 55', respectively. Find the distance AB. 

2. In a field ABCD, the sides AB, BC, CD, and DA 

are 16, 23, 18, and 29 rods, respectively, and the diagonal 
AC is 34 rods. Find the area of the field. 

3. From the top of a cliff the angles of depression of 
two stakes in the plain beloAv, in line with the observer, 
and 725 feet apart, are found to be 35° 10' and 19° 40', respec- 
tively. Find the height of the cliff above the plain. 



Solution of Oblique Triangles. 93 

4. The area of a triangle is 437, and two of its sides are 
36 and 43. Find the angle between them. 

5. From a point in the same horizontal plane with the 
base of a tower, the angle of elevation of its top is 42°, and 
from a point 200 feet farther away, it is 26°. Find the 
height of the tower, and the distance of its base from each 
point of observation. 

6. Two vessels start at the same point, at the rates of 
9.7 and 5.5 miles an hour, respectively, the first due east, and 
the second due southwest. Find the distance between them 
at the end of an hour and a half, and the bearing of each 
from the other. 

7. Two sides of a triangle are .85 and .74, and the differ- 
ence between their opposite angles is 18° 27'. Solve the 
triangle. 

8. The area of a triangle ABC is 980, its angle A is 
56° 20', and its side b is 44. Find B, c, and a. 

9. The sides of a triangle are 5, 7, and 9, respectively. 
Find the radius of the inscribed circle. 

(By Geometry, the area of a triangle is equal to one-half its perim- 
eter multiplied by the radius of the inscribed circle.) 

10. Two sides of a parallelogram are 8 and 5, and include 
an angle of 61°. Find the diagonals. 

11. The diagonals of a field ABCD intersect at E at an 
angle of 78°. If AE, BE, CE, and DE are 27, 31, 59, and 
64 feet, respectively, find the area of the field. 

12. The bases of a trapezoid are 49 and 95, and the angles 
at the extremities of the latter are 64° and 71°. Find the 
non-parallel sides. 

13. Two vessels, A and B, are sailing due northeast. At 
a certain time, B lies 8 miles due south of A, and at the 
expiration of an hour 75° east of south. If the rate of A 
is 6 miles an hour, find the rate of B. 



94 Plane Trigonometry. 

14. From a point in the same horizontal plane with the 
base of a tower, the angle of elevation of its top is 39° ; and 
from a point 150 feet vertically above the first, the angle of 
depression of the top is 43°. Find the height of the tower, 
and its distance from the first point of observation. 

15. From two points on either side of, and in line with, a 
tower, 300 feet apart, the angles of elevation of its top are 
observed to be 31° and 27°, respectively. Find the height 
of the tower. 

16. From a point in the same horizontal plane with the 
base of a tower, the angle of elevation of its top is 22°, and 
its bearing 31° west of north. From another point 400 feet 
west of the first, the bearing is 26° east of north. Find the 
height of the tower. 

17. One of the non-parallel sides of a trapezoid is 15, the 
angle between it and the longer base is 78°, the angle at the 
other extremity of the longer base is 62°, and the shorter 
base is 9. Find the other two sides. 

18. Two sides of a parallelogram are 103 and 54, and one 
of the diagonals is 137. Find the angles of the parallelo- 
gram, and the other diagonal. 

19. From a ship, two lighthouses bear due northwest. 
After sailing 18 miles in a direction 35° west of south, the 
lighthouses bear 6° west of north and 9° east of north, re- 
spectively. Find the distance between the lighthouses. 

20. The sides AB and BC, of quadrilateral ABCD, are 9 
and 5, respectively, and the angles. A, B, and C are 84°, 
109°, and 96°, respectively. Find the sides AD and CD. 

21. From a position at the foot of a hill surmounted by a 
tower, the angle of elevation of the top of the tower is 56°. 
After walking 1260 feet toward the foot of the tower, up a 
slope whose angle with a horizontal plane is 29°, the tower 
subtends an angle of 65°. How far is the top of the tower 
above the horizontal plane of the foot of the hill ? 



Solution of Oblique Triangles. 95 

22. The sides AB, BC, and CD, of quadrilateral ABCD, 
are 23, 41, and 36, respectively, and the angles B and C are 
116° and 131°, respectively. Find the side AD and the 
angles A and D. 

23. The diagonals of a parallelogram are 58 and 92, 
respectively, and intersect at an angle of 55°. Find the 
sides and angles of the parallelogram. 

24. From two points on the slope of a hill, in the same 
vertical plane with the summit, the angles of elevation of 
the top are 11° and 18°, respectively. The points are 300 feet 
apart, and the second 40 feet above the horizontal plane of 
the first. How far is the top of the hill above the hori- 
zontal plane of the first point ? 

25. To find the distance between two inaccessible buoys, 
A and B, a line CD, 150 feet in length, is measured on the 
shore. At C the angles ACD and BCD are observed to be 
83° and 69°, respectively, and at D the angles ADC and 
BDG are observed to be 74° and 97°, respectively. Find 
the distance AB. 

26. A bluff, with a lighthouse on its edge, is observed 
from a boat, the angle of elevation of the top of the light- 
house being 25°. After rowing 1000 feet directly toward the 
lighthouse, the angles of elevation of its top and bottom are 
found to be 53° and 39°, respectively. Find the height of 
the bluff, and of the lighthouse. 



SPHERICAL TRIGONOMETRY. 



D>©<C 



IX. GEOMETRICAL PRINCIPLES. 

115. If a triedral angle be formed with its vertex at the 
centre of a sphere, it intercepts on the surface a spherical 
triangle. 

The triangle is bounded by three arcs of great circles, 
called its sides, which measure the face angles of the tri- 
edral angle. 

The angles of the spherical triangle are the spherical 
angles formed by the adjacent sides ; and each is equal to 
the angle between two straight lines drawn, one in the plane 
of each of its sides, perpendicular to the intersection of these 
planes at the same point. 

The sides of a spherical triangle are usually expressed in 
degrees. 

116. A spherical triangle is called right when it has a 
right angle ; quadrantal when it has one side a quadrant. 

117. Spherical Trigonometry treats of the trigonometric 
relations between the sides and angles of a spherical tri- 
angle. 

The face and diedral angles of the triedral angle are not 
altered by varying the radius of the sphere ; and hence the 
relations between the sides and angles of a spherical triangle 
are independent of the length of the radius. 

97 



98 spherical Trigonometry. 

118. We shall limit ourselves in the present work to such 
triangles as are considered in Geometry, where each angle 
is less than two right angles, and each side less than the 
semi-circumference of a great circle; that is, where each 
element is less than 180°. 

119. The proofs of the following properties of spherical 
triangles may be found in any treatise on Solid Geometry : 

1. Any side of a spherical triangle is less than the sum 
of the other two sides. 

2. The sum of the sides of a spherical triangle is less 
than 360°. 

3. The sura of the angles of a spherical triangle is greater 
than 180°, and less than 540°. 

4. If A'B'C is the polar triangle of spherical triangle 
ABO, that is, if A, B, and C are poles of sides B'C, CA', 
and A'B', respectively, then ABC is the polar triangle of 
spherical triangle A'B'C. 




5. In two polar triangles, each angle of one is measured 
by the supplement of that side of the other of which it is 
the pole ; that is, 

a' = 180° -A. b' = 180° - B. c' = 180° - C. 

A' = 180° -a. B' = 180° - b. C = 180° - c. 

6. If two angles of a spherical triangle are unequal, the 
sides opposite are unequal, and the greater side lies opposite 
the greater angle ; conversely, if two sides of a spherical 
triangle are unequal, the angles opposite are unequal, and 
the greater angle lies opposite the greater side. 



Geometrical Principles. 99 

120. A spherical triangle is called tri -rectangular wlien it 
has three right angles ; each side is a quadrant, and each 
vertex is the pole of the opposite side. 

121. I. Let O be the right angle of right spherical tri- 
angle ABC, and suppose a < 90° and b < 90°. 




E 

Complete the tri-rectangular triangle A'B'C; also, since 
B' is the pole of AC, and A' of BC, construct the tri- 
rectangular triangles AB'B and A' BE. 

Then since B lies within triangle AB'B, AB or c is < 90°. 

Since BC is < B'C, ZAis<Z B'AD, or < 90°. 
Since ACis < A' C, Z B is < Z A'BE, ot KdO"". 

II. Suppose a < 90° and b > 90°. 




h 

Complete the lune ABA'C. 
Then in right triangle A'BC, A'C= 180° - b. 
That is, sides a and A'C of triangle A'BC are each < 90° ; 
and by I., A'B and angles A' and A'BC are each < 90°. 

But, c = 180° - A'B, A = A', and B = 180° - A'BC. 

Whence, c is > 90°, A < 90°, and B > 90°. 

L.ofC. 



loo spherical Trigonometry. 

Similarly, if a is > 90° and h < 90°, then c is > 90°, 
A > 90°, and B < 90°. 

III. Suppose a > 90" and b > 90°. 




Complete the lune ACBO. 
Then in right triangle ABO, 

AC = 180° - b, and BC = 180° -a. 

That is, sides AC and BC of triangle ABC are each 
< 90° ; and by I., AB and angles BAC and ABC are 
each < 90°. 

But, A = 180° - BAC, and 5 = 180° - ABC. 

Whence, c is < 90°, A > 90°, and B > 90°. 

Hence, in any right spherical triangle : 

1. If the sides about the right angle are in the saine quad- 
7'ant, the hypotemise is < 90° ; if they are in differeiit quad- 
rants, the hypotenuse is > 90°. 

2. An angle is in the same quadrant as its opposite side. 

122. In the figure of § 119, we have, by § 119, 1, 
a' <b' -\- c'. 

Putting for a', b', and c' the values given in § 119, 5, we 
have 

180°-^<180°-5 + 180°- O, or 5+ O-^<180°. 

Again, by § 118, B -\- C + 180° > A. 

Whence, B+C-Ai8>- 180°. 

Therefore, ^ + C - ^ is between 180° and - 180°. 

Similarly, C+A — B and ^ + ^ — C are between 180° 
and - 180°. 



Right Spherical Triangles. 



lOI 



X. RIGHT SPHERICAL TRIANGLES. 

123. Let G be the right angle of right spherical triangle 
ABC, and the centre of the sphere. 




Draw radii OA, OB, and OC. 

At any point A of OA, draw lines J. '5' and A'C in 
planes GAB and OAC, respectively, perpendicular to OA, 
meeting OB and OC at J5' and C\ respectively ; also, draw 
line B'C. 

Then, OA is perpendicular to plane A'B'C. 

Whence, each of the planes A'B'C and OBC is perpen- 
dicular to plane OAC, and hence B'C is perpendicular 
to OAC. 

Therefore, B'C is perpendicular to A'C and OC 

By § 115, sides a, b, and c measure angles BOC, COA, 
and AOB, respectively, and the angle A of the spherical 
triangle is equal to angle B'A'C. 

In right triangle OA'B', we have 

.,r,r„ OA' OC OA' 
.osc = .osA'OB' = ^ = —X^r 

But in right triangles OB'C and OCA', 



Whence, 



OC . OA' , 

= cos a, and = cos o. 

OB' OC 

cos c = COS a cos b. 



(69) 



I02 Spherical Trigonometry. 

B'C 



And, cos^ = cos5'^'0'=-jy^, = ^y^ = — — • (71) 



In like manner, sin B = — — ? (72) 



and cos 5 = ^^^^^. (73) 

tan c 



B'C 


OB' 
'A'B' 


sin a 


A'B'~ 


sm c 




OB' 






A'C 




A'C 
A'B'~ 


OA' 

' A'B' 

OA' 


tan& 
tan c 


sin 5 






smc 






tana 







(70) 



124. From (70) and (71), we obtain 



, J sin A sin a tan c sin a 

tan ^ = = — X 



cos A sin c tan b cos c tan b 



Whence by (69), tan A = ^^L^^ = ^^^^^ (74) 

cos a cos b tan o sm b 

In like manner, tan 5 = -; (75) 

sm a 



125. By (4), sin a = cos a tan a ; then (70) may be written 

tana 

cos a tan a tan c 

sm ^ = = 

cos c tan c cos c 

cos a 
Whence by (69) and (73), 

A COS 5 /_-s 

sm ^ = (76) 

cos & 

In like manner, sin B = • (77) 

cos a 



Right Spherical Triangles. 103 

126. From (69), (76), and (77), we have 

7 cos A cos B 1. A 1. T> /„rt\ 

cos c = cos a cos = -: X = cot A cot B. (78) 

sm ^ sm ^ 

127. The proofs of § 123 cannot be regarded as general, 
for in the construction of the figure we have assumed a and 
6, and therefore c and A (§ 121), to be less than 90°. 

To prove formulae (69) to (73) universally, we must con- 
sider two additional cases : 

Case I. WJien 07ie of the sides a and b is < 90°, and the 
other > 90°. 

c^ B 




In right spherical triangle ABC, let a be < 90° and b > 90°. 
Complete the lune ABA'C; then, in spherical triangle 
A'BC, 

A'B=lSO°-c, A'C=lSO°-b, A'=A, and A'BC=1S0°-B. 

But by § 121, c is > 90°, A < 90°, and B > 90°. 
Hence, each element, except the right angle, of right 
spherical triangle A'BC is < 90° ; and we have by § 123, 

cos A'B = cos a cos A'C, 

,inA' = -^^, sin A'BC = '^4^, 

sm A'B' smA'B' 

tSLnA'B' tan ^'5 

Putting for A'B, A'C, A' and A'BC their values, we have 
cos (180° - c) = cos a cos (180° - 6), 



I04 Spherical Trigonometry. 

sin A = . '^^^ . sin (180° - B)= sin (180° -6) 

sin(180°-cy ; ^ sin (180° -cy 

cos A = ^^^(180°-^) ^^g .^g()o _ ^x tana 

tan (180° -c) ^ ^ tan (180° -c) 

Whence, by § 32, — cos c = cos a (— cos b), 



sin ^ 


sma 
sin c 


cos A 


— tan 6 

— tan c 



cosB 



sm c 

tan g 
— tanc 



and we obtain formiilse (69) to (73) as before. 

In like manner, the formulae may be proved to hold when 
a is > 90° and b < 90°. 

Case II. When both a and b are > 90°. 

In right spherical triangle ABO, let a and 6 be > 90°. 
Complete the lune ACBC. 
By § 121, c is < 90°, A > 90°, and B > 90°. 
Hence, each element, except the right angle, of right 
spherical triangle ABG' is <90°; and we have by § 123, 




cosc = 


cos 


AC 


cosBC, 




smc 






sin ABC 


sin AC 
smc 


tanc 






Gos ABO' 


tan 5(7' 
tanc 



Right Spherical Triangles. 105 

Putting for AC, BC, BAG' and ABC their values, 
cos e = cos (180° - a) cos (180° - b), 

sin(180°-^) = ^HL(l§l^^^^, 



smc 



^ ^ tano 



sin(180°- 


B) = 


sin (180°- 
sine 


-*). 




cos(180°- 


.B) = 


tan (180°- 
tanc 


-a). 




ence, by § 32, cos c = (- 


- cos a) ( — 


cos 


b), 


A sin a 

sm A = , 

smc 




sin 


L^:. 


_ sin 6 
smc 


. — tan b 

— cos ^ = , 

tanc 




— cos 


B = 


— tana 
tan c 



and we obtain formulae (69) to (73), as before. 

128. The formulge of §§ 123 to 126 are collected below 
for convenience of reference : 

cos c = cos a cos b. 



sin^-^i"". 


sinB-^!^''. 


smc 


smc 


A tan 5 

cos^ = 

tanc 


^^^ -p tan a 

cos B = 

tanc 


tan^-*^"". 


, ^ J. tan 6 
tanjt> =-: 


sm& 


sma 


si^4_cosB 


ginJJ_COS^, 


coso 


COS a 


cosc = 


= cot A cot B. 




io6 Spherical Trigonometry. 

The student should compare the formulae for the sines, 
cosines, and tangents of A and B with the corresponding 
formulae in §§ 2 and 4. 

129. liTapier's Rules of Circular Parts. 

These are two rules which include all the formulae of 
§ 128. 

CO. B 



CO. A 



In any right spherical triangle, the elements a and b, and 
the complements of elements A, B, and c (written in abbre- 
viated form, CO. A, co. B, and co. c), are called the circular 
parts. 

If we suppose them arranged in the order in which the 
letters occur in the triangle, any one of the five may be 
taken and called the middle part; the two immediately 
adjacent are called the adjacent parts, and the remaining 
two the opposite parts. 

Then Napier's rules are : 

I. The sine of the middle part is equal to the product of 
the tangents of the adjacent p)arts. 

XI. The sine of the middle part is equal to the product of 
the cosines of the opposite parts. 

130. Napier's rules may be proved by taking each cir- 
cular part in succession as the middle part, and showing 
that the results agree with the formulae of § 128. 

1. If a be taken as the middle part, h and co. B are the 
adjacent parts, and co. c and co. A the opposite parts. 
Then the rules give 
sin a = tan h tan (co. B), and sin a = cos (co. c) cos (co. A), 



Right Spherical Triangles. 107 

Or by § 31, sin a = tan h cot B, and sin a ~ sin c sin A ; 
which are equivalent to (75) and (70). 

2. If h be taken as the middle part, a and co. A are the 
adjacent parts, and co. c and co. B the opposite parts. 

Then, sin b = tan a tan (co. A) = tan « cot A, 

and sin 6 = cos (co. c) cos (co. B) = sin c sin B : 

which are equivalent to (74) and (72). 

3. If CO. c be taken as the middle part, co. A and co. B 
are the adjacent parts, and a and b the opposite parts. 

Then, 

sin (co. c) = tan (co. A) tan (co. B), and sin (co. c) = cos a cos 6. 

Or, cos c = cot A cot 5, and cos c = cos a cos b ; 
which agree with (78) and (69). 

4. If CO. A be taken as the middle part, b and co. c are 
the adjacent parts, and a and co. B the opposite parts. 

Then, 
sin (co. ^) = tan b tan (co. c), and sin (co. A) = cos a cos (co. B). 

Or, cos ^ = tan b cot c, and cos A = cos a sin jB ; 
which are equivalent to (71) and (77). 

5. If CO. B be taken as the middle part, a and co. c are 
the adjacent parts, and b and co. A the opposite parts. 

Then, 
sin(co. 5) = tan a tan (co. c), and sin(co. 5) = cos b cos (co. A). 

Or, cos -B = tan a cot c, and cos B = cos 5 sin A ; 
which are equivalent to (73) and (76). 

Writers on Trigonometry differ as to the practical value of 
Kapier's rules ; but in the opinion of the highest authorities, 
it seems to be regarded as preferable to attempt to remem- 
ber the formulae by comparing them with the analogous 
formulae for plane right triangles, as stated in § 128. 



io8 Spherical Trigonometry. 

SOLUTION OF RIGHT SPHERICAL TRIANGLES. 

131. To solve a right spherical triangle, two elements 
must be given in addition to the right angle. 

There may be six cases : 

1. Given the hypotenuse and an adjaceyit angle. 

2. Given an ajigle and its opposite side. 

3. Given an angle and its adjacent side. 

4. Given the hypotenuse and another side. 

5. Given the tivo sides a and b. 

6. Given the two angles A and B. 

132. Either of these cases may be solved by aid of § 128. 
If any two elements are given, the formula for computing 

either remaining element may be found as follows : 

Take the formula which involves the given parts and the 
required part. 

If all the remaining elements are required, the following 
rule will be found convenient : 

Take the three forniulm ivhich involve the given parts. 

133. It is convenient to have a check on the logarithmic 
work, which may always be done without the necessity of 
looking out any new logarithms. 

Examples of this will be found in § 136. 
The check formula for any particular case may be selected 
from the set in § 128 by the following rule : 

Take the formida which involves the three required parts. 

Note. If Napier's rules are used, the following rule will indicate 
which of the circular parts corresponding to the given elements and 
any required element is to be regarded as the middle part. 



Right Spherical Triangles. 109 

If these three circular parts are adjacent, take the middle one as the 
middle part, and the others are then adjacent parts. 

If they are not adjacent, take the part which is not adjacent to 
either of the others as the middle part, and the others are then oppo- 
site parts. 

For the check formula, proceed as above with the circular parts 
corresponding to the three required elements. 

Thus, if c and A are the given elements, 

1. To find a, consider the circular parts a, co. c, and co. A ; of 
these, a is the middle part, and co. c and co. A are opposite parts. 
Then, by Napier's rules, 

sin a = cos (co. c) cos (co. A) = sin c sin A. 

2. To find b, the circular parts are b, co. c, and co. A ; in this case 
CO. A is the middle part, and b and co. c are adjacent parts. Then, 

sin (co. A) = tan b tan (co. c), or cos A = tan b cot c. 

3. To find B, the circular parts are co. B, co. c, and co. A; co. c is 
the middle part, and co. A and co. B are adjacent parts. Then, 

sin (co. c) = tan (co. A) tan (co. B), or cos c = cot A cot B. 

4. For the check formula, the circular parts are a, b, and co. B ; 
a is the middle part, and b and co. B are adjacent parts. Then, 

sin a = tan b tan (co. B) = tan b cot B. 

134. In solving spherical triangles, careful attention 
must be paid to the algebraic signs of the functions ; the 
cosines, tangents, and cotangents of angles between 90° and 
180° being taken negative (§ 21). 

It is convenient to place the sign of each function just 
above or below it, as shown in the examples of § 136 ; the 
sign of the function in the first member being then de- 
termined in accordance Avith the principle that, in multipli- 
cation or division^ like signs produce +, and unlike signs 
produce — . 

Note. In the examples after the first of § 136, the signs are 
omitted in every case where both functions in the second member 
are positive. 



no spherical Trigonometry. 

135. In finding angles corresponding, if the function is a 
cosine, tangent, or cotangent, its sign determines whetlier 
the angle is acute or obtuse ; that is, if it is +, the angle is 
acute; and if it is — , the angle is obtuse, and the supple- 
ment of the acute angle obtained from the tables must be 
taken (§ 32). 

If the function is a sine, since the sine of an angle is 
equal to the sine of its supplement (§ 32), both the acute 
angle obtained from the tables and its supplement must 
be retained as solutions, unless the ambiguity can be 
removed by the principles of § 121. 

EXAMPLES. 

136. 1. Given 5=z33°50', a ==108°; find A, b, and c. 

By the rule of § 132, the formulae from § 128 are, 

. T) cos J. +„ -D tan h „^„ t> tan a 

sm B = ^ tan B = ■? cos B — 

cos a sin a tan c 

— — + + + + — +Q rj a 

That is, cos A — cos a sin B^ tan 6 = sin a tan 5, tan c = 

cos^ 
+ 
Hence, log cos A = log cos a + log sin B. 

log tan h = log sin a + log tan B. 

log tan c = log tan a — log cos B. 

Since cos A and tan c are negative, the supplements of the acute 
angles obtained from the tables must be taken (§ 135). 

Note 1. When the supplement of the angle obtained from the 
tables is to be taken, it is convenient to v^rite 180° minus the element 
in the first member, as shown below in the cases of A and c. 

By the rule of § 133, the check formula for this case is 

cos A = ^^ , or log cos A = log tan h — log tan c. 
tan c 

The values of log tan h and log tan c may be taken from the first 
part of the work, and their difference should be equal to the result 
previously found for log cos A. 



Right Spherical Triangles. 1 1 1 

log COS a = 9.4900 - 10 log tan a = 0.4882 

log sin 5 = 9. 7457 - 10 log cos 5 = 9. 9194 - 10 

logcos^ = 9.2357 - 10 log tan c =: 0.5688 

180° -^ = 80° 5.5'. 180° - c = 74° 53.8'. 

^ = 99° 54.5'. 0=105° 6.2'. 

log sin a = 9.9782 - 10 Check. 

log tan B = 9.8263 - 10 log tan h = 9.8045 - 10 

log tan h = 9.8045 - 10 log tan c = 0.5688 



b = 32° 31. 1'. log cos A = 9.2357 - 10 

2. Given c = 70° 30', ^ = 100°; find a, h, and ^. 

In this case, the three formulae are, 

sin^=^HIi!, cos^=^^^, cos c = cot ^ cot 5. 
sin c tan c 

+ - - + - 

That is, sin a = sin c sin A^ tan h = tan c cos A^ cot B = cos c tan A. 

Here, the side a is determined from its sine ; but the ambiguity is 
removed by the principles of § 121 ; for a and A must be in the same 
quadrant. Therefore a is obtuse, and the supplement of the angle 
obtained from the table must be taken. 

By § 133, the check formula is 

tan B = ^^^ , or sin a = tan b cot B. 
sin a 

Note 2. The check formula should always be expressed in terms 
of the functions used in determining the required parts ; thus, in the 
case above, the check formula is transformed so as to involve cot B 
instead of tan B. 

log sin c = 9.9743 - 10 log cos c = 9.5235 - 10 

log sin ^ = 9. 9934 - 1 log tan A = 0.7537 

log sin a = 9.9677 - 10 log cot B = 0.2772 

180° - a = 68° 10'. 180° - B ^ 27° 50.6'. 

a = 111° 50'. B = 152° 9.4'. 

log tan c =0.4509 

log cos A = 9.2397 - 10 Check. 

log tan b = 9.6906 - 10 log tan 6 = 9.6906 - 10 

180° - 6 = 26° 7.5'. log cot B = 0.2772 

& = 153° 52.5'. logsin a =9.9678-10 



112 Spherical Trigonometry. 

Note 3. We observe here a difference of .0001 in the two values 
of log sin a. This does not necessarily indicate an error in the work, 
for such a small difference might easily be due to the fact that the 
logarithms are only approximately correct to the fourth decimal place. 

3. Given a = 132° 6', h = 77° 51' ; find A, B, and c. 
In this case, the three formulse are, 

^ ~ A tan a , -r, tan b ~ ~ '^ , 

tan A = : tan B = » cos c = cos a cos b. 

sin b sin a 

+ 

The check formula is 

cos c = cot A cot B, or cos c tan A tan B = 1. 

That is, log cos c + log tan A + log tan B = log 1 = 0. _ , 

log tan a = 0.0440 log cos a = 9.8263 - 10 

log sin b = 9.9901 - 10 log cos b = 9.3232 - 10 

log tan ^ = 0. 0539 log cos c = 9. 1495 - 10 

180° - ^ = 48° 32.8'. 180° - c = 81° 53.4'. 

A = 131° 27.2' c = 98° 6.Q'. 

log tan b =0.6670 Check. 

log sin a = 9.8704 - 10 log cos c = 9.1495 - 10 

log tan ^ = 0. 7966 log tan A = 0.0539 

B = 80° 55.4'. log tan ^ = 0. 7966 

log 1 = 0.0000 

4. Given A = 105° 59', a = 128° 33' ; find b, B, and c. 
The formulae are, 

smb=^^^, g.+^_cos^ 



tan A cos a sin A 

The check formula is sin B = 

sin c 

In this example, each required part is determined from its sine ; 
and as the ambiguity cannot be removed by § 121, both the acute 
angle obtained from the tables and its supplement must be retained in 
each case. 



Right Spherical Triangles. 113 

log tan a = 0.0986 log sin a = 9.8932 - 10 

log tan ^ = 0.5430 log sin ^ == 9.9828 - 10 

log sin b = 9.5556 - 10 log sin c = 9.9104 - 10 
b = 21° 3.9', c =54° 26.7', 

or 158° 56.1'. or 125° 33.3'. 

iogcosJ[ = 9.4399 -10 ^^^^^^ 

log cos a = 9.7946 - 10 ^^^ ^^ ^ ^ ^^^^g _ ^^ 

log sin B = 9.6453 - 10 ^^^ ^.^ ^ ^ 9.9104-10 

B = 26° 13. 5', j^g g.^ ^ ^ g g^g^ _ ^Q 

or 153° 46.5'. 

It does not follow, however, that these values can be combined pro- 
miscuously ; for by § 121, since a is > 90°, with the value of b less 
than 90° must be taken the value of c greater than 90°, and the value 
of B less than 90° ; while with the value of b greater than 90° must be 
taken the value of c less than 90°, and the value of B greater than 90°. 

Thus the only solutions are : 

1. 6 = 21° 3.9', c = 125° 33.3', _B = 26° 13.5'. 

2. 6 = 158° 56.1', c = 54° 26.7', 5 = 15-3° 46.5'. 

Note 4. The figure shows geometrically why there are two solu- 
tions in this case. 

B.... 



For if AB and ^0 be produced to A', forming lune ABA'C, tri- 
angle A'BC has side a and angle A' equal, respectively, to side a and 
angle A of triangle ABC^ and both triangles are right-angled at C. 

It is evident that sides A'B and A'C and angle A'BC are the sup- 
plements of sides c and b and angle ABC, respectively. 

Solve the following right spherical triangles : 

5. Given a = 31°, c = 60°. 

6. Given A = 27°, B = 73°. 



114 Spherical Trigonometry. 

7. Given a = S% ' 6 = 22°. 

8. Given B = 25°, c = 34°. 

9. Given A = 40°, a = 26°. 
10. Given 5 = 127° 20', a = 82°. 

- 11. Given b = 18°, c = 112° 10'. 

12. Given A = 120°, c = 161° 50'. 

13. Given A = 159° 40', b = 135°. 

14. Given 5 = 110° 50', 5 = 118° 30'. 

15. Given a = 49° 10', 6 = 100°. 

16. Given A = 170° 50', b = 55°. 

17. Given A = 28° 20', c = 108° 40'. 

18. Given A = 104° 50', B = 156° 30'. 

19. Given a = 164° 10', c = 133°50'. 

20. Given 5 = 99° 40', c = 50°30'. 

21. Given b = 130° 40', c = 70° 10'. 

22. Given a = 129° 30', 6 = 166° 50'. 

23. Given ^ = 24° 31', 6 = 19° 9'. 

24. Given A = 83° 15', a = 76° 46'. 

25. Given B = 115° 22', a = 145° 39'. 

26. Given 6 = 43° 57', c = 62°5'. 

27. Given A = 81° 29', B = 131° 51'. 

28. Given a = 147° 35', c = 52°13'. 

29. Given A = 139° 4', c = 63° 47'. 

30. Given B = 39° 43', a = 54° 26'. 

31. Given b = 153° 18', c = 121° 54'. 

32. Given A = 37° m\ b = 157° 12'. 

33. Given B = 114° 38', c = 168° 23'. 



Right Spherical Triangles. 115 

34. Given a = 66° 6', c- 109° 44'. 

35. Given A = 30° 48', c = 13° 27'. 

36. Given B = 69° 16', a = 160° 55'. 

37. Given a = 142° 42', 5 = 78° 6'. 

38. Given A = 126° 53', B = 47° 34'. 

39. Given B = 16° 24', c = 140° 37'. 

40. Given B = 98° 17', b = 143° 8'. 

137. Quadrantal Triangles. 

By § 119, 5, tlie polar triangle of a quadrantal triangle is 
a right spherical triangle. 

Hence, to solve a quadrantal triangle, we have only to 
solve its polar triangle, and take the siqjplements of the 
results. 

1. Given c = 90°, a = 67° 38', 6 = 48° 50'; find A, B, 
and C. 

Denoting the polar triangle by A'B'C, we have by § 119, 5, 

C = 90°, A' = 112° 22', B' = 131° 10' ; to find a', 6', and c'. 

By § 132, the formulae for the solution are 

cos a' = ^2^, cos b' = £2^, and cos c' = co't A' co"t B'. 
sin B' sin A' 

+ + 

The check formula is cose' = cos a' cos 6'. 

log cos A' = 9.5804 - 10 log cot A' = 9.6143 - 10 

log sin B' = 9.8767 - 10 log cot B' = 9.9417 - 10 

log cos a' = 9.7037 - 10 log cos c' = 9.5560 - 10 
180° -a' = 59° 38.2'. c' = 68° 54.8'. 

logcos^' = 9.8184 -10 Check. 

log sin A = 9.9660 - 10 log cos a' = 9.7037 - 10 

log cos b' = 9.8524-10 ^^^ ^°« ^' = ^'^^^^ " ^^ 

180° - b' = 44° 36.7'. log cos c' = 9.5561 — 10 



ii6 Spherical Trigonometry. 

Then in the given quadrantal triangle, we have 
A = 180°-a'= 59° 38.2', 
^ = 180° -6'= 44° 36.7', 
(7=180°-c' = lll° 6.2'. 

EXAMPLES. 

Solve the following quadrantal triangles : 

2. Given ^ = 157°, (7=121°. 

3. Given a = 117°, b = 142° 50'. 

4. Given ^ = 43°, J5 = 106°. 

5. Given & = 162°20', (7= 64° 40'. 

6. Given ^ = 30° 10', a =72° 30'. 

7. Given ^ = 118° 16', 6 =137° 57'. 

8. Given a = 51° 34', C=25°49'= 

9. Given 5 = 141° 13', 0=49° 35', 

10. Given a = 17° 41', B = 38° 24'. 

11. Given B = 159° 2', b = 136° 28'. 

138. Isosceles Spherical Triangles. 

We know, by Geometry, that if an arc of a great circle be 
drawn from the vertex of an isosceles spherical triangle to 
the middle point of the base, it is perpendicular to the base, 
bisects the vertical angle, and divides the triangle into two 
symmetrical right spherical triangles. 

By solving one of these, we can find the required parts 
of the given triangle. 

1. Given a = 115°, 6 = 115°, C=71°40'; find A, B, 
and a 

Denoting the elements of one of the right triangles by A', B', C, 
a', 6', and c', where C is the right angle, we have 

c' = a = 115°, and A' = iC = 35° 50'. 



Right Spherical Triangles. 117 

We have then to find the parts a and B in this triangle. 

By § 128, sin A^ = ?HL«!, and cos d = cot A' cot B'. 
sin c' 

■ - - + 

Or, sin a' = sin c' sin A', and cot B' = cos c' tan A'. 

log sin c' = 9.9573 - 10 log cos c' = 9.6259 - 10 

log sin A' = 9.7675 - 10 log tan A' = 9.8586 - 10 

log sin a' = 9. 7248 - 10 log cot B' = 9.4845 - 10 

a' = 32° 3.0'. 180° - B' = 73° 1.8 . 

B' = 106° 58.2'. 
Then in the given isosceles triangle, 

A = B=B' = 106° 58.2', and c = 2 a' = 64° 6.0'. 

EXAMPLES. 

Solve the following isosceles spherical triangles . 

2. Given A = 27°, B = 27°, c = 135°. 

3. Given a =. 152°, 6=152°, 0= 68°. 

4. Given a = 112° 36', b = 112° 36', c = 123° 50', 

5. Given ^ = 159° 14', 5 = 159° 14', a =137° 47'. 



ii8 



Spherical Trigonometry. 



XL OBLIQUE SPHERICAL TRIANGLES. 

GENERAL PROPERTIES OF SPHERICAL TRIANGLES. 

139. In any spherical triangle, the sines of the sides are 
proportional to the sines of their opposite angles. 





:.:.....-'D 



Let ABC be any spherical triangle, and draw arc CD per- 
pendicular to AB. 

There will be two cases according as CD falls upon AB 
(Fig. 1), or AB produced (Fig. 2). 

In right spherical triangle ACD, in either figure, we have 
by (70), 

sinCi) 



Also, in Fig. 1, 
And in Fig. 2, 



sin J. = 
sin 5 



sin 6 
sin(7i) 



sma 
sin5=sin(180°-(75i)) 
= sin CBD (§ 32) = 



sin CD 



sma 



Dividing these equations, we have in either case 

sin (72) 
sin A sin h sin a 



sin B sin CD sin h 
sin a 



(79) 



Oblique Spherical Triangles. 119 

T T1 sin^ sin 5 . . 

In like manner, -. — 77 = - — j (80) 

sm C sm c ^ ^ 

- sin^ sina , ,. 

and -. — 7^ = -. • (81) 

sm C sm c ^ 



140. In any spherical triangle, the cosine of any side is 
equal to the product of the cosines of the other tivo sides, plus 
the continued product of their sines and the cosine of their 
included angle. 

In right spherical triangle BCD, in Fig. 1, § 139, we have, 
by (69), 

cos a = cos BD cos CD = cos (c — AD) cos CD. 
And in Fig. 2, 

cos a = cos BD cos CD = cos (AD — c) cos CD. 
Then in either case, by (12), 

cos a = cos c cos AD cos CD + sin c sin AD cos CD. 

But in right spherical triangle ACD, by (69), 

cos AD cos CD = cos h. 

cos & 

Also, sin AD cos CD = sin AD 7^ = cos h tan J.i). 

' cos AD 

-D 4- • ^ T. sin5 sin 5 

But, smce tan = =-, cos b = - — -• 

cos b tan b 

tan AD 
Then, sin AD cos Oi) = sin b — — r— = sin b cos ^, by (71). 

tan o 

Whence, cos a = cos & cos c + sin 6 sin c cos A. (82) 

In like manner, 

cos b = cos c cos a + sin c sin a cos B, (83) 

and cos c = cos a cos b + sin a sin b cos C (84) 



I20 Spherical Trigonometry. 

141. Let ABC and A'B'C be a pair of polar triangles. 




Applying formula (82) to triangle A'B'C, we obtain 

cos a' = cos b' cos c' -\- sin b' sin c' cos A'. 
Putting for a', b', c', and A' the values given in § 119, 5, 
cos (180° -A) = cos (180° - B) cos (180° - C) 

-f sin (180° - B) sin (180° - C) cos (180° - a). 
Whence, by § 32, 

— cos A = (— cos B) (— cos (7) + sin B sin C(— cos a). 
That is, 

cos A = — cos 5 cos + sin B sin (7 cos a. (85) 
Similarly, 

cos B = — cos (7 cos yl + sin C sin ^ cos b, (86) 

and cos C = — cos J. cos ^ + sin ^ sin B cos c. (87) 

The above proof illustrates a very important application 
of the theory of polar triangles in Spherical Trigonometry. 

If any relation has been found between the elements of a 
spherical triangle, an analogous relation may be derived 
from it, in which each side or angle is replaced by the 
opposite angle or side, with suitable modifications in the 
algebraic signs. 

142. To express the sines, cosines, and tangents of the half- 
angles of a spherical triangle in terms of the sides of the 
triangle. 



Oblique Spherical Triangles. 121 

From (82), § 140, 

sin h sin c cos ^ = cos a — cos & cos c. 

-rxTi A cos a — cos h cos c . .. 

Wnence, cos A = : -, (A) 

sm b sm c 

Subtracting both members from 1, we have 

^ A ^ cos a — cos 6 cos c 

1 — cos A = l : : 

sm b sm c 

_ cos b cos c + sin b sin c — cos a 
sin 6 sin c 
Whence, by (28), 

2 gin^ ' A = ^^^ (^ ~ ^) ~ ^^^ ^ 

^ sin b sin c 

But by (20), 

cos y — cos x = 2 sin |- (a? + 2/) sin ^ (^ — V)- (B) 

Whence, 

2 gjn^ 1 ,1 ^ ^ ^^"^ ^ ^^^ + (P-^)1 s^^ i [^ - (^ - c) ] ^ 
^ sin ?> sin c 

. „ , . sin i (a -1- 5 — c) sin i (a — 6 + c) 

or sm^ ^A = ^^ -. — i . -- 

^ sm sm c 

Denoting the sum of the sides, a-\-b ^ c^ by 2 s, we have 

a + &-c = (a + 64-c)-2c = 2s-2c = 2(s-c), 

and a-6 + c = (a + 6 + c)-2& = 2s-2& = 2(s- 6). 

sin is — b) sin (s — c) 



Whence, sin- 



sin b sin c 



Or, sm|^=J«i'iii^4^i!lii^ 

' ^ ^^ sm o sm c 



(88) 



T T1 • 1 T^ /sm (s — c) sm (s — a) ^-.v 

In like manner, sm i J5 = V ^ — - - -^ (Q^) 

^ ^ sm n sm a 



sm c sm a 



. . -, ^ /sin (s — a) sin (s — b) , . 

and sm i = \ ^— - — —^1 (90) 

^ ^ sm a sm h ^ ^ 



122 Spherical Trigonometry. 

Again, adding both, members of (A) to 1, we have 
^ , A -i , cos a — cos b cos c 

1 + cos ^ = 1 -I : : ■ 

sm b sm c 

_ cos a — (cos b cos c — sin 6 sin c) 
~~ sin 6 sin c 

Whence, by (29), 

^ n , . cos a — cos (6 + c) 
2cosn^ = 



2 



sin 6 sin c 



2 sm 4 f& + c + a) sm 1 f & + c — a) , ,_,x 

= ^-^ ■• — r^ — ^-^ -5 by (B). 

sm sm c ./ v / 

Putting a-\-b-\-c = 2s, whence 5 + c — a = 2 (s — a), 

„ , . sin s sin (s — a) 
^ sm sm c 



^ , . /sin s sin (s — a) , . 

Or, cos i ^ = ^--^-^-\-—-l. (91) 



sin 6 sin c 



T ,., , ^ /sin s sin (s — 6) , . 

In like manner, cos -^ B ='\/ — ^.^^ ^ ^^^^ ^^ — -, (92) 



sm c sm a 



1 ^ /sin s sin (s — c) . . 

and cos i O = \ = . ^ ^ ' (93) 

^ ^ sm a sm 6 ^ ^ 

Dividing (88) by (91), we have 



in i ^ /sin (s — 5) sin (s — c) / sin & sin c 
OS -J- J. ~ \ sin 5 sin c ^' sin s sin (s — a) 



sm 
cos 



-r-rr, . 1 A /Sin (S — b) SlTl (S — C) , ^ 

Whence, tan i ^ = V ^ ■ r ^ ' (9*) 



sin s sin (s — a) 



In like manner, tan IB= \ ^ / "'^^^ — —, (95) 

' ^ ^f smssm(.s-&) ^ ^ 

and tani(7=JgHL(lz :^)«;n(^-^) . (gg) 

i( sin .s sm (Si — rA 



sin s sm (s — c) 



Oblique Spherical Triangles. 123 

143. To express the sines, cosines, and tangents of the half- 
sides of a spherical triangle in terms of the angles of the 
triangle. 

From (85), § 141, sin B sin C cos a = cos ^ -f cos 5 cos G. 

Whence, cos a = 22^ii+^2i^oosC. ^^^ 

sin B sin C ^' 

Then, l_eosa = l-22iA±^5^BcosO. 

sin B sin C 

Ov 9 oiT^2 1 ^ _ — (cos S cos — sin B sin O) — cos A 

yji, L sm 2^ a — : — — — -. — — 

sm B sm G 

^ cos (^ + C) + COS A 
sin 5 sin (7 
Then by (19), 

2cosi(^+0 + ^)cosi(^ + C-^) 

i^/ bill -s- (X/ — ; — ; 7^ • 

^ sm B sm C 

Denoting the sum of the angles, A-\- B -\- C, by 2 /S, we 
have 5 + O - ^ = 2 (aS - ^). 

Whence, sin^ \a = - ^^^^^^^(.S - ^) , 

sin S sin (7 



c\^ • 1 / cos aS cos iS — A) ,^„. 

Or, sm 1 a = A^ ■ -d - n (^^) 

\ sm S sm C 

T Ti • 1 r / COS S cos (>S' — -B) /^_x 

In like manner, sm ^ 6 =\ ^ — ^ ^. — '-, (98) 

>/ sm 6 sm ^ 

-, . ^ \ cos /S cos (aS — C) /rto\ 

and smic=:\ -. — -A — =; — '-' (99) 

^ sm J[ sm 5 

Again, adding both members of (A) to 1, we have 

cos^ 4- cos B cos C 



1 H- cos a = 1 4- 



sin 5 sin C 



cos ^ + cos B cos + sin B sin C 
sin B sin C 



124 Spherical Trigonometry. 

Then, 2 cos^ i a = cos ^ + cos (ij - C) 
sin B sin (7 

^ 2cos|[^ + ^-0]cosi[^-(^-C)] 
sin B sin (7 

Or, cosHc» = °"^*(-^ + ^-^)'^"^t(-^-^+^). 

"^ sm i? sm (7 

Y^vA A+ B - C =2{S - G), 2.nd. A- B + C =2{8 - B). 

Whence, cos^ i a^ c os (^ - ^) cos (^ - 0) , 
sm ^ sm (7 



Or, cosi«=xp(^^^)''°^(f-^X 

^ sm 5 sm (7. 

In like manner, 



(100) 



^ \ sin r/sin /I ^ ^ 



sin C sin ^ 
and 



cos J ^ >os(>S-^)cos(^-i?) . (102) 



sin ^ sin B 
Dividing (97) by (100), we have 



tania=J cos >S cos (.S - ^) ^ . 

2 \ COS (/S - j5) cos (/^ - (7) ^ ^ 

In like manner. 



, 1 , / COS S cos (^ — 5) /-.r./iN 

*"^ *' =V- eos(^-a)cos(^-^) - ^''*) 

and -■ tanic=JIZ^p^M^^. (105) 

^ >/ cos (^ - ^) cos (/S - ^) ^ 

NAPIER'S ANALOGIES. 

144. Dividing (94) by (95), we have 



tan iA_ /sin (s — b) sin (s — c) / sin s sin (s — b) 
tan ^ 5 ~~ ^Z sin s sin (s — a) ^ sin (s — c) sin (s — a) 



„ sin ^ ^ cos jB _ /sin^ (^ — ^) _ sin (s — &) 

' cos 4 ^ sin -1- ^ ~ ^Z sin^ (s — a)~ sin (s — a) 



Oblique Spherical Triangles. 125 

Whence bj composition and division, 

sin i A cos ^B -\- cos ^ A sin \ B _ sin (s — b) -{- sin (s — a) 
sin 1 J. cos i 5 — cos -L yl sin i ^ ~~ sin (s — 5) — sin (s — a) 

Then by (9), (U), and (21), 

sin {lA-{-^B) _ tan i [s — & + s — a] 
sin (i^ - i jB) ~ tani^[s - 6 - (s - a)]* 

But s — 6 + 8 — rt = 2s — a — 6 = c. 

^„, sini(^ + ^) tanic 

AMience, ^ — f-^-j ^ = z — rv^ — t^' (^06) 

' sm ^(A — B) tan i (a — 6; ^ ^ 

145. Multiplying (94) by (95), Tve have 



tan i ^ tan i B= / sin (s-b) sin (.9- c) / sin (s-c) sin (s- a) 
^' sin s sin (s — cO ^ sin s sin (s — b) 



{s — «) ^ sin s sin (s — b) 
sin -L ^ sin ^ B _ /sin- (s — c) _ sin (s — c) 



Or _ . _ 

' cos ^ ^4 cos -J- -B ^^ sin- s sin 



Whence by composition and division, 

cos ^ A cos Y ^ — sin ^ A sin ^ 1? _ sin s — sin (s — e) 
cos ^ ^ cos ^ -B + sin ^ ^ sin -i- 5 sin s + sin (s — c) 

Or, by (21), 

cos(i^4-i^) _ tani[g-(s-c)] 
cos (^A — ^B)~ tan |- [s 4- s — c] * 

But s-\-s — c = 2s — c = a-\-b. 

TTT-L cos-i-(^ + ^) tanic ,,^„. 

Whence, , ) a -ni = i — w \ .. - (107) 

cos ^(^ — 5) tan i (a + 6) 

146. Applying formula (106) to triangle A'B'C, in the 
figure of § 141, we obtain 

sin-i(^^ + ^') _ tanic' 
sin i (A' - £') ~ tan i (a' - 6') ' 



126 spherical Trigonometry. 

But, 

i{A' + B') = ^(180° - a 4- 180° - 6)= 180° - -i.(a + 6); 
^(^' _ 5')- 1(180° -a- 180° -i-b) = -^(a-b); 
1 c' = i(180° - 0)= 90° - 1 (7; 
and i(a' - 6')= J(180° ~A- 180° + 5) = _ i (^4 - J5). 
Whence, 

sin[180°-i(a + &)] _ tan (90° -^C) 
sin[- i-(a - 5)] "" tan [- i-(^ _ 5)]' 

Therefore, by §§ 28, 31, and 32, 

sin i (a + 5) _ cot ^ C 



— sin i- (a — b) — tan -i- ( J. — ^) 

sin|(a + 6) ^ cotiC ^ , 

' sin i (a -6) tan 1(^-5) ^ ^ 

In like manner, from (107), we obtain 

cosi(^'4-5')_ tanic' • 



cos i (A' - 5') tan i (a' + 5') 
But, 

1 (a' + 5')= ^(180° - ^ + 180° - 5)= 180° - ^(^ + 5). 

Whence, 

cos [180° - ^ (g + 6)] ^ tan (90° -jC) 
cos [ - 1 (a - 6)] ~ tan [180° - ^ (^ + 5)] * 

Therefore, by §§ 28, 31, and 32, 

— cos |-(a + &) _ cot ^(7 

cos^(a — 6) ~" — tan i- (^ + 5) ' 

cosi(a + 6) _ cot -1(7 ^ 

cos I (a -6) tan K^ 4- 5^ ^ 



Oblique Spherical Triangles. 127 

147. The foriniilse exemplified in §§ 144, 145, and 146 
are known as Napier^ s Analogies. In each case there may 
be other forms according as other elements are used. 



SOLUTION OF OBLIQUE SPHERICAL TRIANGLES. 

148. In the solution of oblique spherical triangles, we 
may distinguish six cases : 

1. Given a side and the adjacent angles. 

2. Given two sides and their included angle. 

3. Given the three sides. 

4. Given the three angles. 

5. Given two sides and the angle opposite to one of them. 

6. Given two angles and the side opposite to one of them. 

By application of the principles of § 119, 5, the solution 
of an example under Case 2, 4, or 6, may be made to depend 
upon the solution of an example under Case 1, 3, or 5, 
respectively ; and vice versa. 

Hence, it is not essential to consider more than three cases 
in the solution of oblique spherical triangles. 

The student must carefully bear in mind the remarks 
made in §§ 134 and 135. 

149. Case I. Given a side and the adjacent angles. 

1. Given A = 70°, B = 131° 20', c = 116°; find a, b, and O. 

By Napier's Analogies (§§ 144, 145), we have 
sm^(B-\-A)_ tan|c cos^(_S + J.)_ tanic 



sini(^ — ^) tan 1(6— a) cosl(B—A) tan 1(6 + a) 

Whence, tan ^ (6 — a) = sin ^(B — A) esc ^ (_B + J.) tan ic, 

- + - + 

and tan ^ (6 + a) = cos \{B — A) sec \{B + A) tan i c. 



128 spherical Trigonometry. 

From the data, l(B- A)= 30° 40', l(B+ A)= 100° 40', ^ c = 58°. 

log sin 1 (^ - ^) = 9.7076 - 10 log cos i ( 5 - ^) = 9.9346 - 10 

log CSC 1{B + A)= 0.0076 log sec ^B + A) = 0.7326 

log tan 1 c = 0. 2042 log tan i c = 0. 2042 

log tan K& - «) = 9.9194 - 10 log tan K& + «) = 0.8714 

1(6 -a) =39° 42.8'. 180° - i (6 + a)= 82°20.5'. 

!(& + «) =97° 39.5'. 
Then, « :^ i (5 + «) _ i (6 _ «)=: 57° 56.7', 

and ■ 5 3^ 1 (6 + a) + 1 (6 _ a) = 137° 22.3'. 

To find C, we have by § 146, 

^ . ^ sin i(6 + a)^ ^ .^ ._ 
cot 1 O = ■ ,,^ — ^tan UB - A) 

^ sm 1 (6 - a) ^ ^ ^ - ' 

= sini(6 + a) esc |(& — a) tani(5 — A). 

log sin 1 (6 + «) = 9.9961 - 10 
log CSC i (6 - a) =0.1946 
logtani(5-Yl)= 9.7730- 10 
logcoti(7 = 9.9637 - 10 

1(7 = 47° 23.6', and C= 94° 47.2'. 
Note 1. The value of C may also be determined by the formula 

coti(7=:-^^f4^^tani(5 + ^) (§146). 
^ cos 1(6 — a) ^^ ^ ^'' ^ 

Note 2. The triangle is alv^ays possible for any values of the 
given elements. 

EXAMPLES. 

Solve the following spherical triangles : 

2. Given A = 87°, B = 61°, c = 112°. 

3. Given 5 = 41°, 0=122°, a = 37°. 

4. Given A = 135°, O = 51°, b = 69°. 



5. Given ^ = 147° 30', B = 163° 10', c = 76° 20 

(For additional examples under Case I., see § 155.) 



Oblique Spherical Triangles. 129 

150. Case II. Given two sides and their included angle. 
1. Given 6=: 137° 20', c=116°, ^-70°; find 5, O, and a. 
By Napier's Analogies (§ 146), we have 
sini(6 + c)_ cot^^ cosi(6 + c)_ coti^ 



sini(&-c) tani(S-C) cosi(6-c) tan 1(5+0) 

Whence, tan \{B — C) = sin |(6 — c) esc ^(6 + c) cot 1 A, 

- + - + 

and tan \{B + C)= cos l(b — c) sec J (& + c) cot 1 A. 

From the data, 1 (6 - c) = 10° 40', 1 (6 + c) = 126° 40', ^A = 35°. 

log sin 1(6 - c) = 9.2674 - 10 log cos i(& -c) = 9.9924 - 10 

log esc 1(6 + c) = 0.0958 log sec J (6 + c) = 0.2239 

log cot 1^=0. 1548 log coti^ = 0.1548 



logtan^(J?- (7) =9.5180 -10 logtani(5+ 0=0.3711 

K-B- 0)= 18° 14.5'. 180° -1(5+ = 66° 57.1' 

1(5+ C)= 113° 2.9'. 

Then, 5 = i(^ + C') + K^ - <^)= 131° 17.4', 

and C=l(B+ C)-U^- C)= 94° 48.4'. 

To find a, we have by § 144, 

tania = ^Hlii^^t_^tani(&-c). 
' sin 1(5-0) '^ ^ 

log sin i(^ + C) = 9.9639 - 10 
log CSC K^- 0)= 0.5044 
log tan i(& - c) = 9.2750 - 10 
logtania = 9.7433 - 10 

la = 28° 58.3', and a =57° 56.6'. 
Note. The triangle is possible for any values of the given elements, 

EXAMPLES. 

Solve the following spherical triangles : 

2. Given a = 64°, /) := 34°, C = 48°. 

3. Given 6 = 42°, c = 96°, ^ = 110°. 



ijo Spherical Trigonometry. 

4. Given a = 146°, c = 69°, B = 125°. 

5. Given a = 90° 50', 6 = 117° 50', (7=120°. 
(For additional examples under Case II., see § 155.) 

151. Case III. Given the three sides. 

The angles may be calculated by the formnlse of § 142. 

If all the angles are to be computed, the tangent formulae 
are the most convenient, since only four different angles 
occur in the second members. 

If but one angle is required, the cosine formula involves 
the least work. 

The triangle is possible for any values of the data, pro- 
vided that no side is greater than the sum of the other two, 
and that the sum of the sides is less than 360° (§ 119, 1 
and 2). 

If all the angles are required, and the tangent formulae 
are used, it is convenient to modify them as follows. 

By (91), 



tan i J. = \ r^^ ^^ ~ ^^ ^^^ ^^ ~ ^^ ^^" ^^ ~ ^^ 
^ ^^ sin s sin^ (s — a) 



1 / sin (s — a) sin (s — b) sin (s —c) 

sin (s — a) ^ sin s 



Denoting J ^m (^ - «) sm (;^ - &) sin (. - c) ^ ^^ ^^ ^^^^ 

\ Sin .Q 



tan ^A = 



sm 



sin (s — a) 



k k 

Similarly, tan IB = ^—, — — — , and tan \C= —. — 7 -• 

•^ ^ sm (s — 6) ^ sm (s — c) 

1. Given a = 57°, b = 137°, c = 116° ; find A, B, and C. 

Here, 2 s = a + b + c = S\0°. 

Whence, s = 155°, s~ a = 98°, s-b = 18°, s-c = 39°. 



Oblique Spherical Triangles. 131 

log sin(s - a) = 9.9958 - 10 log k = 9.8294 - 10 

log sin(s - 6) = 9.4900 - 10 log sin(s -b)= 9.4900 - 10 

log sin (s - c)= 9.7989- 10 

log CSC s = 0.3741 



log 


tan 1 5 = 


0.3394 






i^ = 


66° 24.2'. 






B = 


130° 48.4 


/ 




\ogk = 


9.8294 - 


10 


log sin (s - c) = 


9.7989 - 


10 


log 


tan 1 C = 


0.0305 






^c = 


47" 0.8'. 






C = 


94° 1.6'. 





2 )19.6588 - 20 

logA; = 9.8294- 10 

log sin(s ~ a)= 9.9958 - 10 

log tan 1^ = 9.8336 - 10 

1^ = 34° 17.0'. 

^ = 68° 34.0'. 



EXAMPLES. 
Solve the following spherical triangles : 

2. Given a = 69°, 6 = 74°, c = 63°. 

3. Given a = 103°, 6 = 53°, c = 61°. 

4. Given a = 91°, b = li8°, c = 132°. 

5. Given a = o8°, 6 = 138°, c = 116°; find A. 
(For additional examples under Case III., see § 155.) 

152. Case IV. Given the three angles. 

The sides may be calculated by the formulse of § 143. 

If all the sides are to be computed, the tangent formulae 
are the most convenient, since only four different angles 
occur in the second members. 

If but one angle is required, the sine formula involves the 
least work. 

The triangle is possible for any values of the data, pro- 
vided that the sum of the angles is between 180° and 540° 
(§ 119, 3), and that each of the quantities B -\- C — A, 
C + A-B, and A-\-B-C is between 180° and -180° 
(§ 122). 

For such values of the angles, S is between 90° and 270°, 
and each of the quantities S —A, S —B, and S — C between 
90° and - 90°. 



132 Spherical Trigonometry. 

Then, cos jS is —, while the cosines of S — 'A, S — B, and 
^ _ O are + (§ 21). 

Hence, the expressions under the radical signs in the 
formulae are essentially positive, and no attention need be 
paid to the algebraic signs. 

If all the sides are required, and the tangent formulae are 
used, it is convenient to modify them as follows : 

By (103), 



I _ I cos S cos^ (S — A) 



/Qf A\ I COS xS 

= COS (/iS — ^) a/ - 



COS (^ - ^) COS (>S - 5) COS (^ - 0) 



^'""''^^ ^- cos(S-A)J(S-B)cos(8-C) ^^ ^' 

tan \a = /i cos {S — A). 
In like manner, 
tan \h = Kgos {S — B), and tan \c — Kcos (S — C). 

1. Given A = 150'', B = lSr, (7=115°; find a, b, and c. 

Here, 2S = A + B+ C = 396°. 

Whence, S = 198°, S- A = 48°, S- B = 67°, S-C= 83°. 

log cos aS'= 9.9782 - 10 log ^=0.7375 

log sec (/S'-^)= 0.1745 lOg cos (S -B)= 9.5919 - 10 

log sec (^ - J5) = 0.4081 log tan 1 & = 0.3294 

log sec (a9-0)= 0.9141 ^6 = 64° 54.2'. 

2)JaU9__ ?> = 129° 48.4'. • 

log K = 0. 7375 loo- K = 0. 7375 

log cos (/S-^)^ 9.8255 - 10 log cos{S- C)= 9.0859 - 10 

log tan 1 a = 0. 5630 log tan 1 c = 9.8234 - 10 

la = 74° 42.2'. ic = 33°39.6'. 

a = 149° 24.4'. c = 67° 19.2'. 



Oblique Spherical Triangles. 133 

Note 1. By § 84, cos 198° = - sin 108° = - cos 18° ; whence, with- 
out regard to algebraic sign, log cos 198° = log cos 18°. 

2. Given ^ = 123°, 5-45°, 0=58°; find' a. 



By (97) sin 1 a = V- ''' ^ ''' ^ ^' '^^ • 

' sm B sm C 

Here, 

2S=A-^B-h C= 226° ; whence, S = 113°, and ^S' - ^ = - 10°. 
log cos ^r= 9.5919-10 
log cos ( aS' - ^) = 9. 9934 - 10 
log CSC ^= 0.1505 
log CSC C = 0.0716 

2 )19.8074-20 
log sin I a = 9.9037 - 10 

1 a = 53° 14.4', and a = 106° 28.8'. 
Note 2. By § 28, cos (- 10°) = cos 10°. 



EXAMPLES. 

Solve the following spherical triangles : 

3. Given A = 52°, B = 59°, C = 83°. 

4. Given A =■- 143°, B = 28°, C = 32°. 

5. Given J[ = 142°, 5 = 159°, 0=133°. 

6. Given A = 70°, B = 122°, O = 95° ; find b, 
(For additional examples under Case IV., see § 155.) 

153. Case V. Given two sides and the angle opposite to 
one of them. 

1. Given a = 58°, h = 138°, B = 134° 50' ; find A, C, and c. 

T. /r.«N sin ^ sin a . ^ . 7 . -r> 

By (79) , - — - = - — -, or sm A = sina esc b sni B. 
^ sm B sin& 

log sin a = 9.9284 - 10 

log CSC &= 0.1745 

log sin B = 9.8507 - 10 

logsin^ = 9.9536 - 10 

A = 63° 58.6' or 116° 1.4' (§ 135). 



134 Spherical Trigonometry. 

To find C and c, we have by §§ 144 and 146, 

cot J O = sin ^(6 + a) esc J(6 - a) tan \{B — A), 
and tan J c = sin ^(B + ^4) esc ^(B ~ A) tan J(6 — a). 

Using the first value of A^ 

^{B -{-A)= Q9° 24.3', and ^(B -A) = 35° 25.7^ 
Also, ^{b -j-a)= 98°, and ^(b -a)= 40°. 

log sin iib + a)= 9.9958 - 10 log sin ^(B + A)= 9.9941 - 10 
log esc K& - a) = 0.1919 log esc ^ (J5 - ^) = 0. 2368 

log tan ^(B - A)= 9.8521 - 10 log tan §(& - a) = 9.9238 - 10 

log cot ^ O = 0.0398 log tan ^ c =: 0. 1547 

J (7= 42° 22. 7'. ^c = 54° 59.6'. 

(7 =84° 45.4'. c = 109° 59.2'. 

Using the second value of A, 

^{B + A)= 125° 25.7', and ^{B - ^) = 9° 24.3'. 

log sin J(5 + a) =9.9958- 10 log sin ^(5 + ^)r= 9.9111 - 10 

log CSC ^(b -a)= 0.1919 log esc ^{B - A)= 0.7867 

log tan ^{B - A)= 9.2192 - 10 log tan ^{b - a) = 9.9238 - 10 

log cot J O = 9.4069 - 10 logtanjc = 0.6216 

J a = 75° 40.9'. Jc = 76°33.6'. 

0=151° 21. 8'. c = 153° 7.2'. 

Thus the two solutions are : 

1. ^ = 63° 58.6', C = 84° 45.4', c = 109° 59.2'. 

2. J. = 116° 1.4', C= 151° 21. 8', c = 153° 7.2'. 

As in the corresponding case in the solution of plane 
oblique triangles (compare §§ 108 and 109), there may 
sometimes be two solutions, sometimes only one, and some- 
times none, in an example under Case Y. 

After the two values of A have been obtained, the num- 
ber of solutions may be determined by inspection ; for, by 
§ 119, 6, if a is < 6, ^ must be < 5 ; and if a is > 6, ^ must 
be > 5. 



Oblique Spherical Triangles. 135 

Hence, only those values of A can he retained which are 
greater or less than B according ccs a is greater or less than b. 

Thus, in Ex. 1, a is given < b ; and since both values of 
A are < B, we have two solutions. 

Again, if the data are such as to make log sin ^ positive, 
there will be no solution corresponding. 

2. Given a = 58°, c = 116°, 0=94° 50'; find A 

In this case, ^^^ = ^^^, or sin ^ = sin a esc c sin C. 

sin C sin c 

log sin a =9.9284 - 10 

log CSC c =0.0463 

logsin 0=9.9985- 10 

logsin^ = 9.9732 -10 

^ = 70° 5.0' or 109° 55.0'. 

Since a is given < c, only values of A which are < C can be re- 
tained ; then the only solution is J. = 70° 5.0'. 

3. Given b = 126°, c = 70°, 5 = 57° ; find C. 

In this case, 5i5_^ = ^^5L5, or sin C = sin c esc b sin B. 

sin B sin b 

log sine =9.9730- 10 
log CSC & =0.0920 
logsin5 = 9.9236- 10 



log sin (7 = 9.9886 - 10 

(7= 76° 56.7' or 103° 3.3'. 

Since both values of C are > B, while c is given < &, there is no 
solution. 

EXAMPLES. 

Solve the following spherical triangles : 

4. Given a = 29°, 6 = 14°, ^ = 49°. 

5. Given a = 98°, c = 36°, C = 163°. 



136 Spherical Trigonometry. 

6. Given 6 = 132°, c=:56°, ^ = 116° 18'. 

7. Given a = 104° 50', c = 153°20', ^ = 70°. 

8. Given a = 111° 20', b = 41° 40', B = 25°. 
(For additional examples under Case V., see § 155.) 

154. Case VI. Given tivo angles and the side opposite to 
one of them. 

1. Given ^ = 110°, 5 = 131° 20', 5 = 137° 20'; find a, c, 
and G. 

In this case, ^^^^^ = ^^5^, or sin a = sin ^ esc B sin h. 

sin h sin B 

logsinvl = 9.9730- 10 
logcscJ5 = 0.1244 
log sin & =9.8311 -10 
log sin a =9.9285- 10 

a = 58° 1.2' or 121° 58.8'. 
To find c and C, we have by §§ 144 and 146, 

tanic = sin 1(5+ A) esc |(5 - ^) tan^ (6 - a), 
and cot i C = sin \(h + a) esc \{h - a) tan \{B - A). 

Using the first value of a, 

!(?) + «) = 97° 40.6', and \(h-a)= 39° 39.4'. 
Also, \{B + A)= 120° 40', and \{B-A)= 10° 40'. 

log sin 1{B + A)= 9.9346 - 10 log sin 1 (6 + a) = 9.9961 - 10 

log CSC 1{B-A)^0. 7326 log esc 1 (& - a) = 0. 1951 

log tan 1 (6 - a) = 9.9185 - 10 log tan \{B - A)= 9.2750 - 10 

logtanic = 0.5857 logcoti 0= 9.4662 - 10 

ic = 75°26.9'. 1(7= 73° 41.5'. 

c = 150° 53.8'. 0=147° 23.0'. 

Using the second value of a, 

l(h + a) = 129° 39.4', and \{h - a) = 7° 40.6'. 



Oblique Spherical Triangles. 137 

log sin ^{B + A)= 9.9346 - 10 log sin 1 (6 + «) = 9.8865 - 10 

log CSC i (^ - ^) = 0. 7326 log esc ^ (5 - a) = 0. 8742 

log tan 1 (5 - a) = 9. 1297 - 10 log tan i(B-A)= 9.2750 - 10 

log tan 1 c = 9.7969 - 10 log cot | (7 = 0.0357 

ic = 32=3.9'. 1(7 = 42° 38.8'. 

c = 64°7.8'. = 85° 17.6'. 

Thus the two solutions are : 

1. a = 58° 1.2', c = 150° 53.8', O = 147° 23.0'. 

2. a = 121° 58.8', c = 64° 7.8', C = 85° 17.6'. 

In examples in Case VI., as in Case V., there may some- 
times be two solutions, sometimes only one, and sometimes 
none. 

As in Case Y., only tJiose values of a can be retained which 
are greater or less than b according as A is greater or less 
than B. 

Also, if log sin a is positive, the triangle is impossible. 

EXAMPLES. 
Solve the following spherical triangles : 

2. Given A = 84°, B = 19°, a = 28°. 

3. Given B = 159°, C = 36°, b = 9°. 

4. Given A = 25° 20', C = 153° 27', a = 73° 10'. 

5. Given A = 142° 40', C = 71° 10', c = 39° 30'. 

6. Given A = 110°, B = 123° 20', b = 127°. 
(For additional examples under Case VI., see § 155.) 

MISCELLANEOUS EXAMPLES. 
155. Solve the following spherical triangles : 

1. Given a = 38°, 6 = 51°, c = 42°. 

2. Given B = 116°, C = 80°, c = 83°. 



ijS Spherical Trigonometry. 

3. Given A = 78°, B = 41°, c = 108°, 

4. Given 5 = 99° 40', c = 64°20', ^ = 96° 10'. 

5. Given A = 76°, B = 81°, C = 61°. 

6. Given A = 62% C = 102°, a = 64° 30'. 

7. Given a = 72°, 6 = 47°, (7=33°. 

8. Given ^ = 133° 50', ^=66° 30', a = 81° 10'. 

9. Given a = 101°, 5 = 49°, c = 60°. 

10. Given 5 = 135°, C=50°, a = 70° 20'. 

11. Given a = 162° 20', 6 = 15° 40', 5 = 125°. 

12. Given ^=138° 20', 5 = 31° 10', (7=35° 50'. 

13. Given a = 109° 20', c = 82°, ^ = 107° 40'. 

14. Given A = 132°, B = 140°, 6 = 127^ 

15. Given ct = 60°, c = 98°, 5=110°. 

16. Given a = 55°, c = 138°10', ^ = 42° 30'. 

17. Given ^ = 61° 40', (7 =140° 20', c = 150°20'. 

18. Given a = 61°, 5 = 39°, c = 92°. 

19. Given a = 40°, 5 = 118° 20', ^ = 29° 20'. 

20. Given A = 110°, B = 131°, C = 147°. 

21. Given a = 115° 20', c = 146°20', (7 =141° 10'. 

22. Given B = 73°, C = 81° 20', b = 122° 40'. 

23. Given ^=31° 40', O=122°20', 5 = 40°40'. 

24. Given 5 = 108° 30', c = 40°50', (7 =39° 50'. 

25. Given 6 = 120° 20', c = 70°40', ^ = 50°. 

26. Given B = 22° 20', C = 146° 40', c = 138° 20'. 



Applicarions. 



^39 



XII. APPLICATIONS. 

156. Shortest Distance between Two Points on the Sur- 
face of the Earth. 

In problems concerning navigation, the earth may be 
regarded as a sphere. 

The shortest distance between any two points on the sur- 
face is the arc of a great circle which joins them; the angles 
between this arc and the meridians of the points determine 
the hearings of the points from each other. 




Thus, if Q and R are the points, and PQ and PR their 
meridians, Z. PQR determines the bearing of R from Q, and 
Z PRQ the bearing of Q from R, 

If the latitudes and longitudes of Q and R are known, 
the arc QR and angles PQR and PRQ may be determined 
by the solution of a spherical triangle. 

For if EE' is the equator, and PG the meridian of 
Greenwich, 

Z QPR = Z RPG - Z QPG = longitude R - longitude Q. 

Also, PQ = PE -QE = 90° - latitude Q, 
and PR = PE' + RE' = 90° + latitude R. 

Thus, in spherical triangle PQR, two sides and the in- 
cluded angle are known, and the remaining elements may 
be computed. 



140 Spherical Trigonometry. 

When QB has been found in degrees, its length in miles 
may be calculated by finding its ratio to 360°, and multiply- 
ing the result by the length of the circumference of a great 
circle ; in the following problems, the radius of the earth is 
taken as 3956 miles. 

EXAMPLES. 

157. In each of the following examples, find the shortest 
distance in miles between the places named, and the bearing 
of each from the other : 

1. Havana (lat. 23° 9' N., Ion. 82° 23' W.), and Gibraltar 
(lat. 36° 9' N., Ion. 5° 21' W.). 

2. Batavia (lat. 6° 8' S., Ion. 106° bT E.), and San Fran- 
cisco (lat. 37° 48' N., Ion. 122° 24' W.). 

3. Vera Cruz (lat. 19° 12' N., Ion. 96° 9' W.), and Cape 
of Good Hope (lat. 34° 22' S., Ion. 18° 29' E.). 

4. Auckland (lat. 36° 51' S., Ion. 174° 50' E.), and Callao 
(lat. 12° 4' S., Ion. 77° 13' W.). 

5. Boston lies in lat. 42° 20' N., Ion. 71° 4' W., and Glas- 
gow in lat. 55° 52' N., Ion. 4° 16' W. In what latitude does 
a great circle course from Boston to Glasgow cross the meri- 
dian of 40° W. ? 

6. Yokohama lies in lat. 35° 27' N., Ion. 139° 41' E., and 
Cape Horn in lat. ^^'^ 59' S., Ion. 67° 16' W. In what lati- 
tude does a great circle course from Yokohama to Cape 
Horn cross the meridian of 160° W. ? 

158. The Astronomical Triangle. 

Let be the position of an observer on the surface of 
the earth ; P the celestial north-pole ; Z the zenith. 

The great circle EE' having P for its pole, is called the 
celestial equator; and the great circle HH', having Z for its 
pole, is called the horizon. 



Applications. 



141 



Let S be the position of a star ; PSM a meridian through 
S ; ZSJSf a quadrant of a great circle through Z and S. 

The arc SM is called the declination of the star ; it is 
called declination north or south, according as the star is 
north or south of the celestial equator. 




The angle SPZ is called the hour-angle of the star ; the 
arc SN, its altitude; the angle PZS, its hearing or azimuth. 

The arc EZ is the latitude of the observer. 

The spherical triangle SPZ is called the Astronomical 
Triangle. 

Its sides have the following values : 

SP=PM- jSM= 90° - the declination of the star ; 
SZ = ZN - SN = 90° - the altitude of the star ; 
PZ=EP-EZ = 90° - the latitude of the observer. 

Its angle SPZ is the hour-angle of the star, and its angle 
SZP the azimuth. 

If any three of these five elements are known, the solu- 
tion of a spherical triangle serves to determine the other 
two. 

159. Determination of Longitude and Time. 

If the altitude and declination of the sun are known, and 
the latitude of the observer, the three sides of triangle SPZ 
are known, and the hour-angle SPZ may be computed. 



142 spherical Trigonometry. 

If 24 hours be multiplied by the ratio of this angle to 
360°, we obtain the time required for the sun to move from 
S to the meridian EP. 

If this time be subtracted from 12 o'clock, if the observa- 
tion is made in the morning, or added, if made in the after- 
noon, we obtain the Jiour of the day at the time and place of 
observation. 

If the Greenwich time of the observation be noted on a 
chronometer, the difference between this and the local time 
as calculated above serves to determine the longitude of the 
place of observation. 

In reducing time to longitude, it should be remembered 
that 24 hours of time correspond to 360° of longitude ; that 
is, one hour of time corresponds to 15° of longitude, one 
minute to 15', and one second to 15". 



EXAMPLES. 

160. 1. At a certain place in latitude 40° N., the altitude 
of the sun was found to be 41°. If its declination at the 
time of the observation was 20° N., and the observation was 
made in the morning, how long did it take the sun to reach 
the meridian ? 

2. A mariner observes the altitude of the sun to be 60°, 
its declination at the hour of observation being 6° IST. If 
the latitude of the vessel is 12° S., and the observation is 
made in the morning, find the hour of the day. If the 
observation is taken at 11.40 a.m., Greenwich time, find the 
longitude of the vessel. 

(In this case, the side PZ of the astronomical triangle is 90° plus 
12°.) 

3. At what hour will the sun set in Montreal (lat. 45° 
30' N.), if its declination at sunset is 18° N. ? 

(At sunset, the sun's altitude is 0°, so that the side 8Z of the 
astronomical triangle becomes 90°. ) 



Applications. 143 

4. A mariner observes the altitude of the sun to be 35° 
23', its declination being 10° 48' S. If the latitude of the 
vessel is 26° 13' N., and the observation is made in the 
afternoon, find the hour of the day. If the observation is 
taken at 7.13 p.m., Greenwich time, find the longitude of 
the vessel. 

5. At what hour will the sun rise in Panama (lat. 8° 57' 
N.), if its declination at sunrise is 23° 2' S. ? 

6. What will be the bearing of the sun at 4 p.m. in Kio 
Janeiro (lat. 22° 54' S.), if its declination at that time is 

3° S. ? 

7. What will be the bearing of the sun at sunrise in 
Boston (lat. 42°21'N.), if its declination at that time is 
13°24'K? 

8. What will be the altitude of the sun at 9 a.m. in 
Mexico (lat. 19° 25' N.), if its declination at that time is 
8° 23' N. ? 



144 Plane Trigonometry. 



FORMULAE. 

PLANE TRIGONOMETRY. 

§28. sin(— ^1) = — sin A cos(— ^)= cos J. 

tan ( — ^4) = — tan J.. cot ( — ^4) = — cot A 

sec(— J_)= sec A esc (— J.) = — esc A . 
§ 29. 

sin(90°-h^)= cos A cos (90° + ^) = - sin A 

tan(90° + ^) = - cot A. cot (90° + yl) = - tan A. 

sec (90° + ^)= -CSC A esc (90° + -4)= sec A 
§35. 

1 1 1 

sec a; 



sin X = 


1 

CSC a? 


tan X = — ; — 
cot a; 


cosa^ = 


1 

secic 


cot £C = 7 

tana; 


§36. 




sin X 


cosa; 


§37. 




cos a; 


sma; 


§38. 




sin^a; + cos2a^ = 1. 


§40. 




sec^ a; = 4 + tan^ 
csc^ a^ = 4 + cot^ 



CSC a; 



cos a; 

4 
sin a; 



(1) 



(2) 



(3) 

(4) 

(5) 

(6) 

(') 

(8) 

§41. sin (a; + ?/) = sin a; cos ?/ + cos a^ sin y. (9) 

cos (a; + y) = cos x cos y — sin x sin y. (10) 

§ 43. sin (x — y) = sin x cos y — cos qi sin y. (11) 

cos (x — y) = cos x cos y + sin x sin 2/. (12) 

/ X tan X + tan v /_ _x 

§44 tan(x + ,) = ^-^;^^^^. (13) 

tan X — tan ?/ /, ^ 

^ ^"^ ~ 4 + tan a; tan y 



Formulae. 145 

, , ^ cot X cot y — \- r-, ,-\ 

cot {x^y) = — - — -^ — (15) 

^ ^^ cotiy + cotx ^ 

coti«cot?/ + l , . 

cot (x — y) = — ^^- — (16) 

^ ^^ cot?/ — cot cc ^ 

§45. sin ;K 4- sin ?/ = 2sini(a; + ?/)cos^(a; — ?/). (17) 

sin X — sin 2/ = 2 cos \ {x + /y) sin \ (x — y). (18) 

cos X -f cos ?/ = 2 cos ^ (;c + //) cos ^{x — ?/). (19J 

cos X — cos y = — 2 sin i (a.- + ?/) sin \(x — y). (20) 

g ^^ sin a; + sin y ^ tan|(x + y) /g^x 
sin a.' — sin 1/ tani(a: — y) 

§47. ^ ■ " 

sin 2 a? = 2 sin x cos ic. (22) cos 2 x = cos^ ic — sin^ x. (23) 



cos 2 X = 1 — 2 sin^ x. 


(24) 


cos2a.' = 2cos^a7 — 1. 


(25) 


tan2a.= ^^^^^ • 
l-tan^x' 

§48. 


(26) 


2 cot a; 


(27) 


2siTi^^x=l — coscc. 


(28) 


2cos2-ia.'=l + cosa.\ 


(29) 


, 1 1 — cos X 

tania^=: 

smx 

§97. 


(30) 


cotU-=^+'"''^- 
sm X 


(31) 


4 ^=c- sin 2 A 


(32) 


4.K=G'sm2B, 


(33) 


2/ir=a2cotA 


(34) 


2 K=b'' cot B. 


(35) 


2A^=aHan5. 


(36) 


2K=l)'td.nA. 


(37) 



2 K= a^\c + a){c- a). (38) 2 K= bV(c + b)(c-b). (39) 

2 j5r= ab, (40) 

§ 99. a : 6 = sin ^ : sin B. (41) 

& : c = sin 5 : sin C. (42) 

c : a = sin C : sin A. (43) 



146 Plane Trigonometry. 

^ a-b tanK^-5) ^^ 



6 + _ tan i (5 + 0) 
6-c~tani(^-C)* 



5c 



(45) 



c-a tani((7-^) ^ ^ 

§ 101. a^ ^ 52 + c^ - 2 6c cos A. (47) 

52 ^ c^ + a^ - 2 ca cos J5. (48) 

c2 = a^ + 52 _ 2 ct^ cos a (49) 

§102. cos^^^^y^. ' (50) 

cos jB = — (51) 

2 ca 

cos G = ,, ^ (52) 

2a6 



§103. si„|^=^(i^4^. (53) 



• 1 T. /(s — c) (s — a) . . 

smiB=yl^ ^^ ^. (54) 



sm i O =^^ ^^ ^. (55) 



cosiA=\ \^ ^ - (56) 



cosi5=V^- ^^^) 



-^*^=V^- (^«) 



Formulae. 147 



, , j(s — b)(s — c) , ^ 

tani^ = \^ /^ , ^ ' (59) 



1 -r^ '(s — c)(s — a) , , 



t^-ic=^l^ 



)(s-b) 



s (s — c) 



§124. 

ta 

§125. 



sin-B _ sin 5 
sin C sin c 



(61) 



§ ^04. 

^ -r^ -, . . . , „ ^^ a- sin B sin (7 , , 

2^=:6csinA (62) 2K= ^ — -, (65) 

^ ^ sm^ ^ ^ 

o /' • D /^ox o T^ 6^ sin C sin J. 

2A^casin5. (63) 2/f= -. — (66) 

^ ^ sin5 ^ ^ 

_. ^__ -f • >^ /_ \ r-k TT- c sm .^ sm x> , - 

2K=ab^mC. (64) 2^= ,—^ (67) 

^ ^ sm (7 ^ ^ 



/iT = Vs (s — a) (s — 5) (s — c). (68) 



SPHERICAL TRIGONOMETRY. 

§ 123. cos c = cos a cos 5. (69) 

A sin a /_-,x . -D sin ?> /„„. 

sm ^ = — (70) sm B = (72) 

sm e sm c 

> tan 6 ,__. ^ tana .__^ 

C0SJ. = (71) COS 5 = - (73) 

tan c tan c 



tan^ = *^5^. (74) tan 5 = *^- (75) 

sm h sin a 



. cos 5 x__s . „ cos^ .„„. 

sm^ = (76) sm5 = (77) 

cos 6 cos a 

§ 126. cos c = cot ^4 cot B. (78) 

§139. sjn^^sina, 

sin ^ sm 5 



(80) 



148 Spherical Trigonometry. 

sin C sin c 



sm c sm a 



cos 4-B 

sm c sm a 



(81) 



sm^tL sin a 

§ 140. cos a = cos h cos c -f- sin h sin c cos ^. (82) 

cos h = cos c cos a + sin c sin a cos ^. (83) 

cos c = cos a cos 6 + sin a sin 6 cos 0. (84) 

§141. cos ^ = — cos 5 cos + sin 5 sin (7 cos a. (85) 

cos B = — cos C cos ^1 + sin C sin ^ cos h. (86) 

cos C = — cos A cos 5 + sin A sin 5 cos c. (87) 



§142. sml^^A H^^T^^"^"^""''^ - ' (88) 

\ sin ^ sin a 



• 1 -D sin (s — c) sin (« — a) /__. 

sm 1 B =\\ ^^ — : — ^ . ^ ^- (89) 

^1 sm ^ sin nr, 



. 1 ^ /sm (s — a) sm (s — b) .^_x 

imi(7 = \ ^ — : — ^ . ^ ^- (90) 

\ sm a sm o 



1 . ismssm(s — a) ,„s 

cos i J. = \ : ^^ ^' (91) 

\ sm ^ sm p. 



kin s sin (s-b)^ (92) 

\ sin f^ sin a 



T ^ /sm s sm (s — c) .^^. 

cos i (7 = \ : ^^ — —^' (93) 

^ sm a sm 



^ sm s sm (s — a) 



tan 1 g^ / sm(«-c)sm(s-«) . ^gg^ 

^ sin s sin (s — b) 

tan^C = Jgi^ -".)^'"(^-'') . (96) 

^^ sm s sm (s — c") 



Formulae. 



§143. 



§144. 
§145. 
§146. 



sm 



sm 



1 _ / co s S cos (S — A) ^ 
^^~\ sin^sinO 



J. _ / cos S cos (S — B) 
^ sin (7 sin ^ 



sm^c 



^ / cos >^ cos ( 1^-0) 
^' sin A sin B 



cos 



Xa= cos (^ - B) cos (>S - C) 
^^* \ sin 5 sin (7 



^^^,,^ /c_oK^-01cosl,S^^. 
* sin C sin ^ 



^ sin ^1 sin ^ 



tani 



^=\- 



cos aS cos (S — A) 
'gos(S-B)gos(S- C) 



. i7^_ I COS ^S cos {S — ^) 

^^2 -\/ cos(^-(7)cos(^-^)' 



tan^c 



%(- 



cos /iS' cos {S — 0) 
cos(>S-^)cos(>S- JB)' 



sin \{A-]- B) _ tan ^ c 
sin i (^ — 5) tan \ (a — b) 



cos -i- (^ 


+ B) 


tan i c 


COS ^ (A 


-B) 


tan 1 (a + 6) 


sin -|- (a 


+ ^)_ 


cot i C 


sin i (a 


-5) 


tan i (^ - ^) 


COS -|- (a 


+ &)_ 


cot 10 



cos ^(a— b) tan i (^ + 5) 



149 

(97) 

(98) 
(99) 
(100) 
(101) 
(102) 
(103) 
(104) 
(105) 
(106) 
(107) 
(108) 
(109) 



ANSWERS. 



§ 54 ; pag-e 40. 

13. 85° 56' 37.32". 15. 95° 29' 34.8". 

14. 14° 19' 26.22". 16. 22° 55' 5.952". 

§ 72; pag-e 53. 

2. 1.5441. 6. 2.1673. 10. 2.4592. 14. 3.3434. 

3. 1.4771. 7. 2.3522. 11. 2.8363. 15. 3.8963. 

4. 1.9912. 8. 2.2431. 12. 2.7023. 16. 3.7656. 

5. 1.9242. 9. 2.6232. 13. 2.5741. 17. 4.1494. 

§ 74; page 54. 

2. .1549. 5. 1.5229. 8. .5192. 11. 1.3734. 

3. .2431. 6. .2273. 9. .6478. 12. .8942. 

4. 1.6532. 7. 2.0212. 10. 2.7202. 13. 1.9842. 

§ 77 ; page 55. 

3. 2.4080. 8. 2.2415. 13. .2510. 19. .9132. 

4. .6036. 9. .0954. 14. .4095. 20. .1643. 

5. 1.0485. 10. .1409. 16. .0409. 21. .3726. 

6. 8.1160. 11. .0777. 17. .7264. 22. .1118. 

7. .4704. 12. .3618. 18. .1511. 23. .8618. 

§ 81; page 57. 

2. 0.3801. 5. 8.2831 - 10. 8. 8.3892 - 10. 11. 2.3043. 

3. 1.2252. 6. 7.1303-10. 9. 6.6865-10. 12. 0.1459. 

4. 9.9084 - 10. 7. 3.7693. 10. 9.0124 - 10. 13. 1.6505 



Answers. 



4. 2.8878. 

6. 3.0237. 

6. 8.5177 - 10. 

7. 9.7164 -10. 

8. 1.3028. 

9. 4.9659. 



§ 82 ; pag-e 58. 

10. 7.6055- 10. 18. 1.646. 

11. 6.8560 - 10. 19. 8886. 

12. 0.7144. 20. 545.9. 

13. 3.0155. 21. .01461. 

14. 8.9379 - 10. 22. .003318. 

15. 5.0610 - 10. 23. 102.2. 

30. .00005029. 



24. 


9.493. 


25. 


.2079. 


26. 


44.48. 


27. 


.0001109. 


28. 


63330. 


29. 


.01301. 









§ 87; pages 


61 1 


bo 63. 






1. 


2.151. 


16. 


- .002555. 


31. 


- .3702. 


48. 


.2985. 


2. 


19.38. 


17. 


3692. ^ 


34. 


13.83. 


49. 


.04477. 


3. 


- 3135. 


18. 


.2777. 


35. 


2.487. 


50. 


.7945. 


4. 


.09778. 


19. 


- 15890. 


36. 


1.056. 


51. 


1.805. 


5. 


.009213. 


20. 


.03162. 


37. 


.00002143. 


52. 


179.5. 


6. 


- .1088. 


21. 


244.1. 


38. 


.007105. 


53. 


1.883. 


7. 


6.359. 


22. 


.002791. 


39. 


.6955. 


54. 


- 8894. 


8. 


.03017. 


23. 


.0000002373. 


40. 


.5428. 


55. 


1.344. 


9. 


- 3.119. 


24. 


2.236. 


41. 


- 36.03. 


56. 


- .01335. 


10. 


1327. 


25. 


1.149. 


42. 


- 11.11. 


57. 


37.82. 


11. 


847.8. 


26. 


- 1.220. 


43. 


.9432. 


58. 


.00001146, 


12. 


- .005421. 


27. 


1.778. 


44. 


2.627. 


59. 


.1782. 


13. 


1.205. 


28. 


.6683. 


45. 


2.534. 


60. 


4.698. 


14. 


.2357. 


29. 


.6458. 


46. 


- 1.795. 


61. 


- .03402. 


15. 


- 11.54. 


30. 


.1378. 


47. 


1.032. 







88; page 64. 



9.8556 - 10. 
9.9458 - 10. 
0.6518. 
9.9501 - 10. 
0.8550. 
9.7070 - 10. 
9.7547 - 10. 



8. 9.9535- 

9. 0.7654. 

10. 0.0420. 

11. 83° 5.2'. 

12. 33° 17.8'. 

13. 46° 40.9'. 

14. 73° 33.4'. 



10. 



15. 80° 26.3'. 

16. 31° 20.4'. 



17. 
18. 
19. 
20. 
21. 



8° 53.5'. 
5° 17.6'. 
40° 20.7'. 
66° 43.3'. 
.2960. 



22. 
23. 
24. 
25. 
26. 
27. 



.2482. 
.7033. 

.3886. 
47° 36.3'. 
21° 52.7'. 
49° 57.0'. 
28° 46.7'. 



Answers. 



94; pag-es 67 to 



1. a 



h = 



3. a 

4. A 



12. a 

13. A 
14. 
15. 
16. 



36. 



1.812, & = 6.761. 

12.38, c = 13.35. 

16.78, c = 26.11. 

34° 22.2', b = .5118. 

32° 44.4', c = 49.92. 

10.35, c = 13.14. 

.005916, b = .01269. 

39° 49.1', a = 488.7. 

148.4, c = 948.6. 

49° 55.0'. c = 4.457. 

77.38, c = 91.08. 

3814, b = 3651. 

24° 23.3', a = .02126. 

156.6, c = 856.4. 

.003607, b = .008830 

24840, c = 36090. 

949.8. 34. ^ = 34° 36.7' 

1.491. 37. a = .03446. 



17. 
18. 
19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 



30. 
31. 
32. 



A = 

b = 

.4 = 
a = 
a = 
6 = 

A = 
a = 
b = 
a = 

A = 
a = 
a = 

A = 

a = 



69. 

55° 44.8', c = 4116. 
.6441, c = .6503. 
76° 34.0', a = 2423. 
.2072, b = .4212. 
5091, c = 5268. 
.8478, c = 1.234. 
39° 22.0', & = 121.2. 
8.243, c = 9.275. 
.000005736, c = .00002118. 
.0006772, b = .0003899. 
43° 45.7', 6 = 66650. 
30.51, b = 18.59. 
24540, c = 30010. 
60° 14.1', c=. 007745. 
25.40. 
.2923. 

35. c = 4.488. 



14. 


53° 31.8', 


15. 


135.2 ft. 


16. 


5.036. 


17. 


53° 8.1'. 



§ 96 ; pages 69 to 72. 

2. 416.1 ft. 6. 23.26. 10. 15.27. 

3. 651.8. 7. 285.1 ft. 11. 70.91 ft. 

4. 34.07. 8. 1.235. 12. 121° 0.8'. 

5. 6° 2.3'. 9. 52° 4.2'. 13. 39° 12.0'. 

18. Perimeter, 3.1908; diameter circumscribed circle, 1.2278. 

19. Eadius inscribed circle, 28.58 ; circumscribed, 30.94. 

20. 740.2. 21. Height of cliff, 144.4 ft.; of lighthouse, 153.6 ft. 

22. 1131.3 ft. 25. 17° 1.6'. 28. 14.9 ft. 31. 3° 43.9'. 

23. 62.9 ft. 26. 18.68. 29. 1575 mi. 32. 195.9 ft. 

24. 12°28.9'. 27. 229.02. 30. 109.0 mi. 33. 60.14 ft. 
34. Bearing, S. 42° 28.8' W.; distance, 17.77 mi. 



3.564. 4. 13440. 

.1098. 5. 12.64. 

10. .000001323. 



98 ; page 74. 

6. 46.0. 8. .02036 

7. .0004838. 9. 795. 
11. 2840. 



4 Answers. 

§ 105; pages 82, 83. 

2. 6 = 282.9, c = 268.5. 5. a = 31.49, c = 49.88. 

3. a = 3.384, c = 9.828. 6. « = .5042, 6 =.3618. 

4. a = .02893, 6 = .01825. 7. & = 5499, c = 2959. 

§ 106; pag-e 84. 

2. ^ = 118° 18.0', 6 = 44.73. 5. ^ = 76° 12.9', c = 6.362. 

3. A = 29° 59.5', c = 1419. 6. C = 96° 3.3', 6 = 5141. 

4. (7 = 88° 34.8, a = .4038. 7. 5= 146° 26.3', a = .01044. 

§ 107 ; page 86. 

3. ^ = 44° 25.0', 5 = 78° 28.0', 0=57° 7.4'. 

4. ^ = 71° 47.4', B = 58° 45.6', C = 49° 27.6'. 

5. ^ = 29° 55.2', 5 = 22° 31.4', 0=127° 34.4'. 

6. 71° 33.4'. 7. 31° 6.6'. 8. 134° 29.4'. 

§ 111 ; pag-e 90. 

1. 5 = 48° 32.7', c = 8.522. 6. (7 = 90°, 6 = 5939. 

2. ^ = 29° 21.4', a = 102.2. 7. ^ = 14° 5.4', 6 = .1435. 

3. Impossible. 8. ^ = 25° 31.9', c = 278.4. 

4. C = 44° 56.2', a = 66.68; 9. (7 =46° 21.7', 6 = .8728 ; 
or, (7= 135° 3.8', a = 12.89. or, C= 133° 38.3', 6 = .2351. 

5. Impossible. 

§ 112 ; pages 90, 91. 

1. ^ = 61° 25.7', c= 1018. 2. a = 15.52, 6 = 10.29. 

3. ^ = 44° 37.8', 5 =101° 50.0', (7 = 33° 33.4'. 

4. ^ = 34°25.0', 6 = .7135. 6. (7 =14° 57. 7', 6 = 5074. 

5. ^ = 46° 2.1', a = .05676. 7. a = .0004395, c = .0002092. 

8. ^ = 51° 52.6', 5 = 42° 59.0', (7 = 85° 8.8'. 

9. 5 = 34° 7.8', c = 1223; 11. (7 = 48° 37.3', a = 7.597. 
or, B = 145° 52.2', c = 269.3. 12. 6 = 55610, c = 79270. 
10. (7= 46° 37.8', a =7577. 13. Impossible. 

14. ^ = 54° 5.4', 6 = .01073. 

15. ^=72° 43.8', 5 =47° 40.6', C= 59° 36.6'. 

16. ^ = 90°, c = .1379. 20. ^ = 27°34.2', c = .01875. 

17. Impossible. 21. 0= 134° 36.9', 6 = 273.3. 

18. 6 = .4392, c = .4723. 22. B = 63° 58.6', a = 25660 ; 

19. ^ = 95° 22.2', c = 38.25. or, i5 = n6°1.4', a = 7573. 



Answers^ 5 

§ 113; pages 91, 92. 

2. 582. 7. 24530. 12. 10.28. 17. 7255. 

3. 53.8. 8. .000003186. 13. 883.2. 18. 19.19. 

4. 20.98. 9. .1682. 14. 34840. 19. 47210. 

5. .0780. 10. .0000002941. 15. .03519. 20. .00003759. 

6. 271.3. 11. 28.77. 16. .003042. 21. .000675. 



§ 114; pages 92 to 95. 

1. 608.4 ft. 2. 420.0 sq.rd. 3. 525.8 ft. 4. 34° 22.2' or 145° 37.8'. 

5. Height of tower, 212.8 ft.; distances, 236.4 ft., 436.4 ft. 

6. Distance, 21.20 mi.; bearing of first from second, N. 74° 1.7' E. 

7. Opposite angles, 76° 9.2', 57° 42.2'; remaining side, .6313. 

8. _B = 51°30.5', c= 53.51, « = 46.80. 9. 1.658. 

10. 7.087, 11.30. 11. 3995 sq. ft. 12. 61.51, 58.48. 

13. 14.922 miles an hour. 14. Height, 69.71 ft.; distance, 86.08 ft. 

15. 82.70 ft. 16. 173.2 ft. 17. 19.92, 16.62. 

18. One angle, 118° 6.6'; diagonal, 91.02. 19. 9.012 mi. 

20. ^i> = 9.282, CZ> = 10.65. 21. 1538 ft. 

22. ^i) = 74.98, ^ = 68° 58.3'. 

23. One angle, 59° 44.8' ; sides, 66.99, 37.77. 24. 84.28 ft. 



25. 272.4 ft. 



26. Bluff, 438.7 ft.; lighthouse, 280.1 ft. 



§ 136; pages 113 to 115. 



5. 


^ = 36° 29.4', 


5 = 69° 42.1', 


& = 54° 18.9'. 


6. 


a = 21° 18.0', 


5 = 49° 54.7', 


c = 53°8.2'. 


7. 


^ = 20° 33.8', 


J5 = 70°59.5', 


c = 23° 18.3'. 


8. 


^ = 68° 51.6', 


a = 31° 26.4', 


& = 13° 40.2'. 


9. 


5 = 58° 27.5'. 


6 = 35° 32.4', 


c = 42° 59.3'; 




5 = 121° 32.5', 


6 = 144° 27. 6', 


c = 137° 0.7'. 


10. 


^ = 83° 38.8', 


& = 127° 36.2', 


c = 94° 52.3'. 


11. 


^ = 97° 36.4', 


a = 113° 22.4', 


^=19° 29.4'. 


12. 


a = 164° 20', 


^ = 31° 16.9', 


& = 9°19.1'. 


13. 


a = 165° 18.8', 


B = 104° 13.4', 


c = 46° 50.4'. 


14. 


^ = 48° 10.9', 


a = 44° 29.2', 


c = 109° 52.5' 




^ = 131° 49.1', 


a = 135° 30.8', 


c = 70°7.5'. 









Answers. 






15. 


A = 


49° 35.8', 


B = 


97° 36.0', 


c = 


96° 31.2'. 


16. 


a = 


172° 28.1', 


B = 


84° 45.4', 


c = 


124° 39.3'. 


17. 


a = 


26° 42.8', 


B = 


99° 47.4', 


b = 


110° 59.7'. 


18. 


a = 


129° 66.7', 


b = 


161° 32.5', 


c = 


52° 28.2'. 


19. 


A = 


157° 46.8', 


B = 


74° 12.0 , 


b = 


43° 56.9'. 


20. 


A = 


165° 0.6', 


a = 


168° 29.2', 


b = 


130° 28.2'. 


21. 


A = 


114° 49.3', 


a = 


121° 22.9', 


B = 


126° 14.4'. 


22. 


A = 


100° 38.0', 


B = 


163° 8.0', 


c = 


51° 44.4'. 


23. 


a = 


8° 30. 5', 


B = 


66° 55.5', 


c = 


20° 53.7'. 


24. 


B = 


30° 53.3', 


b = 


30° 12.9', 


c = 


78° 35.0' ; 




B = 


149° 6.7', 


h = 


149° 47.1', 


c = 


101° 25.0'. 


25. 


A = 


138° 15.5', 


b = 


130° 2.3', 


c = 


57° 55.4'. 


26. 


A = 


59° 17.1', 


a = 


49° 26.0', 


B = 


51° 46.0'. 


27. 


a = 


78° 32.1', 


b = 


132° 25.0', 


c = 


97° 42.6'. 


28. 


A = 


137° 17.7', 


B = 


119° 29.5', 


b = 


136° 31.7 . 


29. 


a = 


144° 0.6', 


B = 


110° 57.9', 


b = 


123° 6.1'. 


30. 


A = 


68° 10.6', 


b = 


34° 3.0', 


c = 


61° 11.3'. 


31. 


A=: 


71° 45.5', 


a = 


53° 44.7', 


B = 


148° 2.5'. 


32. 


a = 


16° 48.5', 


B = 


124° 31.6', 


c = 


151° 56.7'. 


33. 


A = 


25° 4.8', 


a = 


4° 53.9', 


b = 


169° 27.2'. 


34. 


A = 


76° 16.7', 


B = 


144° 1.1', 


b = 


146° 26.2'. 


35. 


a = 


6° 50.4', 


B = 


59° 54.0', 


b = 


11° 36.6'. 


36. 


A = 


152° 7.1', 


b = 


40° 48.8', 


c = 


135° 39.6'. 


37. 


A = 


142° 5.8', 


B = 


82° 43.4', 


c = 


99° 26.4'. 


38. 


a = 


144° 24.4', 


b = 


32° 28.8', 


c = 


133° 19.3'. 


39. 


A = 


102° 48.8', 


a = 


141° 46.9', 


b = 


10° 19.1'. 


40. 


A = 


10° 22.5', 


a = 


6° 16.1', 


c = 


142° 41.2'; 




A = 


169° 37.5', 
§ 


a = 
137; 


173° 43.9', 
page 116. 


c = 


37° 18.8'. 


2. 


a = 


152° 52.8', 


b = 


104° 46.7', 


B = 


124° 1.1'. 


3. 


A = 


138° 42.7', 


B = 


153° 25.0', 


C = 


132° 14.3'. 


4. 


a = 


44° 8.1', 


b = 


101° 3.9', 


C^ 


78° 22.3'. 


5. 


a = 


82° 14.4', 


A = 


63° 34.8', 


B = 


164° 4.8'. 


6. 


6 = 


20° 21.2', 


B = 


10° 33.7', 


= 


148° 11.9' ; 




6 = 


159° 38.8', 


B = 


169° 26.3', 


c = 


31° 48.1'. 



Answers. 

7. a = 108° 29.5', 5 =141° 31.9', C= 111° 45.7'. 

8. ^ = 19° 56.3', b = 138° 36.4', B = 163° 15.7'. 

9. « = 46° 49.3', A = 33° 43.3', b = 124° 37.8'. 

10. ^ = 11° 12.1', 6 =76° 13.8', (7 =140° 14.8'. 

11. a = 50° 56.0', ^ = 23° 47.2', (7=31° 17.6'; 
31, a = 129° 4.0', ^ = 156° 12.8', C = 148° 42.4'. 

§ 138; pag-e 117. 

2. C= 159° 59.8', a = 69° 44.6'. 4. (7 = 145° 45.0', ^ = 141° 17.0 

3. ^ = 120° 46.6', c = 30° 26.6'. 5. O = 148° 37.6', c = 80° 36.8'. 

§ 149 ; page 128. 

2. a = 98° 19.9', 5 = 60° 3.9', C= 110° 38.6'. 

3. c = 72° 14.1', 6 = 47° 27.3', ^=32° 24.2'. 

4. a = 120° 34.6', c = 71° 7.6', jB = 50°3.6'. 

5. 6 = 153° 48.4', a = 125° 0.8', C = 140° 24.6'. 

§ 150; pages 129, 130. 

2. A = 110° 46.6', B = 35° 34.2', c = 45° 36.0'. 

3. (7 =78° 65.5', i? = 41°19.7', a = 107° 47.6'. 

4. ^ = 145° 11.8', C = 107° 39.8', b = 126° 37.4'. 

5. jB = 121° 43.2', J. = 105° 52.8', c= 115° 48.6'. 

§ 151; pag-e 131. 

2. ^ = 74° 12.4', J5 = 82°12.0 

3. ^ = 78° 32.0', 5 = 87° 7.0', 

4. ^ = 120° 21.4', 5 = 130° 21.8', (7 

5. 70° 8.8' 

§ 152 ; pag-e 133. 

3. a = 37° 7.2', 6 = 41° 1.6', 

4. a = 101° 34.4', b = 49° 50.4', 

5. a = 125° 16.2', b = 151° 37.4', 

6. 126° 43.4'. 

§ 153 ; pag-es 135, 136. 

4. 5 = 22° 7,4', (7= 112° 17.0', c = 36° 28.0'. 

5. Impossible. 6. C=90°, ^ = 138° 31.6', a = 146° 41.9' 

7. (7= 154° 7.7', J5 = 60°45.8', 5 =63° 52.2'. 

8. ^ = 36° 18.2', (7= 160° 52.8', c = 148° 58.4' ; 
or, A = 143° 41.8', C = 38° 23.0', c = 77° 39.2'. 



c 


= m'' 


'41.4'. 


c 


= 93° 


29.6'. 


c 


= 14C 


>°7.0'. 


c 


= 49° 


28.2'. 


c 


= 59° 


37.2'. 


c 


= 75° 


55.4'. 



8 Answers. 

§ 154; page 137. 

2. 5 = 8° 50.4', c = 27° 37.0', O = 79° 9.2'. 

3. Impossible. 4. c = 90°, B = 8° 14.0', b = 18° 41.2' 

5. a = 155° 56.8', b = 188° 33.6', B = 100° 0'. 

6. a = 63° 55. 7', c = 156°5.8', C = 154° 55.0' ; 
or, a = 116° 4.3' c=72°43.8', = 87° 24.0'. 

§ 155 ; pag-es 137, 138. 



or, 



or, 



or, 



1. 


A =51° 58.8', 


B^ 


= 8.3° 55.2', 


= 58° 53.8'. 


2. 


b = 115° 3.3', 


a - 


= 85° 16.0', 


^ = 81° 24.0'. 


3. 


a = 95° 37. 8', 


& = 


= 41° 52.2', 


= 110° 48.2'. 


4. 


C = 65° 23.3', 


A = 


= 96° 8.0', 


a = 99° 40.0'. 


5. 


a = 68° 25.2', 


b^ 


= 71° 10.8', 


c = 56° 56.8'. , 


6. 


c = 90°, 


B^ 


= 63° 43.4', 


& = 66° 26.8'. 


7. 


^ = 121° 32.8', 


B = 


= 40° 56.8', 


c = 37° 25.8'. 


8. 


Impossible. 








9. 


A = 142° 32.8', 


B = 


= 27° 52.6', 


0=82° 27.2'. 


10. 


b = 120° 16.6', 


c = 


= 69° 19.6', 


^ = 50° 26.2'. 


11. 


Impossible. 








12. 


a = 100° 8.4', 


6 = 


= 50° 1.8', 


c = 60°6.0'. 


13. 


C = 90°, 


B = 


= 113° 36.5', 


6 = 114° 51.9'. 


14. 


a = 67° 24.0', 


= 


= 164° 6.4', 


c = 160° 6.4/ ; 




a = 112° 36.0', 


= 


= 128° 20.6', 


c = 103° 2.4'. 


15. 


C= 86° 59.7', 


^ = 


= 60° 50.9', 


6 = 111° 17.0'. 


16. 


0=146° 37.9', 


jB = 


= 55° 1.2', 


5 = 96° 34.4'. 


17. 


a = 43° 3.1', 


.^ = 


= 129° 8.4', 


i>' = 89°23.8'; 




a = 136° 56.9', 


b-- 


= 20° 35.8', 


5 = 26° 58.6'. 


18. 


^ = 35° 31.0', 


Bz 


= 24° 42.6', 


0=138° 24.8'. 


19. 


5 = 42° 7.9', 


= 


= 160° 12.8', 


c = 153° 37.8'; 




5 = 137° 52.1', 


C-- 


= 49° 38.8', 


c = 89° 51.2'. 


20. 


a = 59° 34.4', 


b-- 


= 136° 10.6', 


c = 150° 1.6'. 


21. 


Impossible. 




22 


, Impossible. 


23. 


c = 64° 19.4', 


a - 


= 34° 3.0', 


5 = 37° 39.6'. 


24. 


5 = 68° 18.0', 


A-- 


= 132° 33.8', 


a = 131° 15.8' i 




B= 111° 42.0', 


A-- 


= 77° 4.6', 


a = 95° 50.0'. 


25. 


5 = 134° 57.3', 


, C-- 


= 50°41.1', 


a = 69° 8.8'. 


26. 


6 = 27° 22.1', 


a - 


= 117° 9.2', 


^ = 47° 21.2'. 



Answers. 9 



§ 157 ; page 140. 

1. Bearing of Havana from Gibraltar, N. 77° 40.3' W. ; of Gibraltar 
from Havana, N. 59" 5.1' E. ; distance, 4593 mi. 

2. Bearing of Batavia from San Francisco, N. 67° 25.5' W. ; of San 
Francisco from Batavia, N. 47° 12.3' E. ; distance, 8650 mi. 

3. Bearing of Cape of Good Hope from Vera Cruz, S. 60° 45.6' E. ; 
of Vera Cruz from Cape of Good Hope, N. 86° 42.4' W. ; distance, 
8330 mi. 

4. Bearing of Callao from Auckland, S. 69° 29.9' E. ; of Auckland 
from Callao, S. 50° 2.5' W. ; distance, 6671 mi. 

5. 54° 35.9' N. 6. 51° 48.0' S. 

§ 160; pages 142, 143, 

1. 3 h. 30 m. 40 s. 

2. Hour of the day, 10 h. 23 m. 33.6 s., a.m. ; longitude, 19° 6.6' W. 

3. 7 h. 17 m. 14.4 s., p.m. 

4. Hour of the day, 2 h. 45 m. 4 s., p.m. ; longitude, 66° 59' W. 

5,. 6 h. 15 m. 21.6 s., a.m. 6. N. 80° 24.6' W. 7. N. 71° 43.7' E. 
8. 45° 5'. 



FOUR PLACE 



LOGARITHMIC TABLES 



TOGETHER WITH A 



TABLE OE NATURAL SINES, COSINES, 
TANGENTS, AND COTANGENTS 



PEEPARED BY 



WEBSTER WELLS, S.B. 

PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY 



BOSTON, U.SA. 

D. C. HEATH & CO., PUBLISHERS 

1900 



Copyright, 1887 and 1900, 
By WEBSTER WELLS. 



TTPOGKAPHY BY J. S. CUSHING & CO., NOEWOOD, MASS. 



Use of the Tables. i 

USE OF THE TABLES. 

I. USE OF THE TABLE OF LOGARITHMS OF NUMBERS. 

This table (pages 12 and 13) gives the mantissse of the 
logarithms of all integers from 100 to 1000, calculated to 
four places of decimals. 

To FIND THE Logarithm of a Number of Three 

Figures. 

Look in the column headed '^ No." for the first two signifi- 
cant figures of the given number. 

Then the required mantissa will be found in the corre- 
sponding horizontal line, in the vertical column headed by 
the third figure of the number. 

Finally, prefix the characteristic in accordance with the 
rules of §§ 66 or 67. 

For example, log 168 = 2.2253; 

log .0344 = 8.5366 - 10 ; etc. 

For a number consisting of one or two significant figures, 
the column headed may be used. , 

Thus, let it be required to find log 83 and log 9. 

By § 80, log 83 has the same mantissa as log 830, and log 
9 the same mantissa as log 900. 

Hence, log 83 = 1.9191, and log 9 = 0.9542.^ 

To FIND THE Logarithm of a Number of more than 
Three Figures. 

Ex. 1. Required the logarithm of 327.6. 
From the table, log 327 = 2.5145, 
and log 328 = 2.5159. 



2 Use of the Tables. 

That is, an increase of one unit in the number produces 
an increase of .0014 in the logarithm. 

Therefore, an increase of .6 of a unit in the number will 
produce an increase of .6 x .0014 in the logarithm, or .0008 
to the nearest fourth decimal place. 

Hence, log 327.6 = 2.5145 + .0008 = 2.5153. 

Note I. The above method is based on the assumption that the 
differences of logarithms are proportional to the differences of their 
corresponding numbers ; which, though not strictly accurate, is suffi- 
ciently exact for practical purposes. 

Note II. The difference between any mantissa in the table and the 
mantissa of the next higher number of three figures, is called the tabu- 
lar difference. The subtraction may be performed mentally. 

The following rule is derived from the above : 

Find from the table the mantissa of the first three significant 
figures^ and the tabular difference. 

Multiply the latter by the remaining figures of the number 
with a decimal point before them. (See Note III.) 

Add the residt to the mantissa of the first three significant 
figures^ and prefix the proper characteristic. 

Note III. In finding the correction to the nearest unit's figure, the 
decimal portion should be omitted provided that, if it is .5 or more 
than . 5, the unit's figure is increased by 1. Thus, 13.26 would be taken 
as 13 ; 30.5 as 31 ; 22.803 as 23. 

Ex. 2. Find the logarithm of .021508. 

Mantissa 215 = 3324 Tabular difference = 21 
2 _^ 

3326 Correction = 1.68 = 2, nearly. 

Result, 8.3326 - 10. 

To FIND THE Number corresponding to a Logarithm. 

Ex. 1. Required the number whose logarithm is 1.6571. 
Find in the table the mantissa 6571. 



Use of the Tables. 3 

In the corresponding line, in the column headed "No.," 
we find 45, the first two figures of the required number, and 
at the head of the column we find 4, the third figure. 

Since the characteristic is 1, there must be two places to 
the left of the decimal point (§ 66). 

Hence, the number corresponding to 1.6571 is 45.4". 

Ex. 2. "Required the number whose logarithm is 2.3934. 

We find in the table the mantissse 3927 and 3945, whose 
corresponding numbers are 247 and 248, respectively. 

That is, an increase of 18 in the mantissa produces an 
increase of one unit in the number corresponding. 

Therefore, an increase of 7 in the mantissa will produce an 
increase of y^g of a unit in the number, or A, nearly. 

Hence, the number corresponding is 247 + .4, or 247.4. 

The following rule is derived from the above : 

Find from the table the next less mantissa, the three figures 
corresponding, and the tabular difference. 

Subtract the next less from the given mantissa, and divide 
the remainder by the tabular difference. (See Note V.) 

Annex the quotient to the first three figures of the number, 
and point off the result. (See Note IV.) 

Note IV. The rules for pointing off are the reverse of the rules 
for characteristic given in ^^QQ and 67. 

1. If —1^ is not written after the mantissa, add 1 to the character- 
istic, giving the number of places to the left of the decimal point. 

2. If —10 is written after the mantissa^ subtract the positive part of 
the characteristic from 9, giving the number of ciphers to be placed 
betiveen the decimal point and first significant figure. 

Ex. 3. Find the number whose logarithm is 8.5265 — 10. 

5265 
Next less mantissa = 5263 ; three figures corresponding, 336. 
Tabular difference = 13)2.00(.15 = .2, nearly. 
13 
70 



4 Use of the Tables. 

By the rule of Note IV. , there will be one cipher between the decimal 
point and first significant figure. 

Hence, the number corresponding = .03362. 

Note V. The correction can usually be depended upon to one 
decimal place ; the division should be carried out to two decimal places 
in order to determine the last figure accurately. (See Note III.) 

II. USE OF THE TABLE OF LOGARITHMIC SINE 
COSINES, ETC. 

This table (pages 14 to 19) gives the logarithms of the 
sines, cosines, tangents, and cotangents of all angles at inter- 
vals of 10 minutes from 0° to 90°. 

For angles between 0° and 45°, the degrees and minutes 
will be found in the left-hand column, and the functions in 
the columns designated by the names at the top; that is, 
sines in the first column, cosines in the second, tangents in 
the third, and cotangents in the fourth. 

For angles between 45° and 90°, the degrees and minutes 
will be found in the right-hand column, and the functions in 
the columns designated by the names at the foot; that is, 
cosines in the first column, sines in the second, cotangents 
in the third, and tangents in the fourth. 

If only the mantissa of the logarithm is found, the charac- 
teristic may be determined from the nearest logarithm in the 
same column in which the characteristic is given. 

Since the sines and cosines of all acute angles, the tan- 
gents of angles between 0° and 45°, and the cotangents of 
angles between 45° and 90°, are less than unity, the charac- 
teristics of their logarithms have been increased by 10, and 
— 10 must be written after the mantissa; in all other cases, 
the true value of the characteristic is given in the table. 

Thus, log sin 38° 30' = 9.7941 - 10 ; 

log tan i55° 20' = 0.3380 ; 
log cot 79° 10' -9.2819 -10; 
log cos 89° 40' = 7.7648 - 10 ; etc. 



Use of the Tables. , 5 

To FIND THE Logarithmic Sine, Cosine, Tangent, or 

Cotangent, of any Acute Angle expressed 

IN Degrees and Minutes. 

Find from the table the logarithmic sine, cosine, tangent, or 
cotangent of the next less m^dtiple of ten minutes, and the 
difference for V corresponding. (See Note VI.) 

Multiply this difference by the number of minutes remaining. 

If sine or tanqent, add ) , . 

^^ . -, \this correction. 

If cosine or cotangent, subtract) 

Note VI. The columns immediately to the right of those headed 
" Sin.," " Cos.," and " Tan.," contain the respective differences for V ; 
the right-hand column of differences is also to be used with the column 
headed "Cot." 

It will be observed that the differences do not stand in the same 
horizontal line with the logarithms, but opposite the intervals between 
consecutive logarithms. For angles below 45°, the difference next he- 
low should be taken ; for angles above 45°, the difference next above. 

Note VII. The rule assumes that the differences of the logarithmic 
functions are proportional to the differences of their corresponding 
angles, which, unless the angle is near to 0° or 90°, is in general suffi- 
ciently exact for practical purposes. (See page 9.) 

Note VIII. If the angle is expressed in degrees, minutes, and 
seconds, the seconds should be reduced to the decimal part of a minute 
before applying the rule. 

Ex. 1. Find Ior tan 17° 14'. 



logtan 17° 10 =9.4898-10 


D. V = 4.5 




18 


4 




Result, 9.4916 - 10 


Corr. = 18.0 




:x. 2. Find log cos 58° 33.5'. 






log cos 58° 30' = 9.7181 - 10 


D. V = 2.1 




7 


3.5 




Result, 9.7174 - 10 


105 
63 






7.35 = 


: 7, nearly 



6 Use of the Tables. 

To FIND THE Acute Angle corresponding to a given 
Logarithmic Sine, Cosine, Tangent, or Cotangent. 

Take from the table, if sine or tangent the next less, if cosine 
or cotangent the next greater, logarithmic function, the angle 
corresponding, and the difference for V. (See Note IX.) 

Find the difference between the given logarithm and that 
taken from the table, and divide it by the difference for V, 
giving the correction in minutes. 

Add the result to the angle corresponding to the next less, or 
next greater, function. 

Note IX. Ill searching for the next less (or greater) logarithm, 
attention must be paid to the fact that the functions are found in 
different columns according as the angle is below or above 45°. 

If, for example, the next less logarithmic sine is found in the column 
with "Sin." at the top^ the angle corresponding must be taken from 
the left-hand column ; but if it is found in the column with " Sin." at 
the foot^ the angle corresponding must be taken from the right-hand 
column. Similar considerations hold with respect to the other three 
functions. 

Ex. 1. Eind the angle whose log sin = 9.9594 — 10. 

9.9594 - 10 
Next less log sin = 9.9590 — 10 ; angle corresponding = 65° 30'. 

D. 1' = .6)4.0(6.66 = 6.7, nearly. 
Adding the correction, the result is 65° 36.7'. 

Ex. 2. Eind the angle whose log cot = 0.1696, 

Next greater log cot = 0.1710 ; angle corresponding = 34° 0' 
0.1696 



•. 1': 


= 2.7)14.0(5.18:: 

13 5 

50 

27 


= 5.2 


, nearly. 






230 




Kesult, 34° 


5.2'. 



Use of the Tables. 7 

To FIND THE Logarithmic Secant or Cosecant of 
ANY Acute Angle. 

Since sec x = and csc x = — — , we have by § 85, 

cos X sm X 

log sec X = colog cos x, and log csc x = colog sin x. 

Hence, to find the logarithmic secant, subtract the logarith- 
mic cosine from 10 — 10; and to find the logarithmic cosecant, 
subtract the logarithmic sine from 10 — 10. 

Ex. Find log sec 22° 38'. 

From the table, log cos 22° 38' = 9.9652 - 10 

Subtracting from 10 — 10, we have 

log sec 22° 38' = 0.0348. 

Note X. The logarithmic cotangent of an angle may be obtained 
by subtracting the logarithmic tangent from 10 — 10. 

To FIND THE Logarithmic Functions op an Angle not 

LYING BETWEEN THE LiMITS 0° AND 90°. 

By § 34, any function of any angle may be expressed as a 
function of a certain acute angle ; and hence the table of the 
functions of acute angles serves to determine the functions 
of angles of any magnitude whatever, positive or negative. 

Ex. Find log sin 152° 16'. 

By § 34, sin 152° 16' = cos 62° 16'. 

Then, log sin 152° 16' = log cos 62^ 16' = 9.6678 - 10. 

Another method would be to find the logarithmic sine of 27° 44', 
the supplement of 152° 16' (§ 32). 

Note XI. If the natural function is negative, as for example in 
the case of the cosine of an angle between 90° and 180°, there is no 
logarithmic function, strictly speaking. (See Note before § 87.) 

In solving examples involving such functions, we proceed as if the 
functions were positive, and determine the algebraic sign of the result 
irrespective of the logarithmic work. Illustrations of this will be found 
in Chapters X. and XI. 



8 Use of the Tables. 

in. USE OF THE TABLE OF NATURAL SINES AND 
COSINES. 

This table (pages 20 and 21) gives the natural vahies of 
the sines and cosines of all angles at intervals of 10 minutes 
from 0° to 90°, calculated to four places of decimals. 

Its use is similar to that of the table of logarithmic func- 
tions, except that the tabular differences for 1' are not given, 
but are to be calculated from the table when required. 

Ex. 1. Find sin 48° 52'. 

The difference between sin 48° 50' and sin 49° 0' is .0019, one-tentli 
of whicli is .00019. 

sin 48° 50' = .7528 D. 1' = 1.9 

4 2 



.7532, A71S. Corr. = 3,8 = 4, nearly. 

Ex. 2. Find the angle whose cos = .5506. 

The difference between the next greater and next less functions, 
.5519 and .5495, is .0024 ; one-tenth of which is .00024. 

Next greater cos = .5519 ; angle corresponding = 56° 30'. 
.5506 



. 1' 


= 2.4)13.0(5.41 z 
12 

100 
96 


= 5.4, 


nearly. 








40 




Result, 


56° 


35.4 



IV. USE OF THE TABLE OF NATURAL TANGENTS 
AND COTANGENTS. 

This table (pages 22 and 23) gives the tangents and 
cotangents of all angles at intervals of 10 minutes from 0° 
to 90° ; its use is similar to that of the table of natural sines 
and cosines. 



Use of the Tables. 9 

V. MORE ACCURATE METHOD FOR FINDING THE 
LOGARITHMIC FUNCTIONS OF ANGLES NEAR 

TO 0° OR 90°. 

It was stated in Note YII., page 5, that in general the 
differences of the logarithmic functions are approximately 
proportional to the differences of their corresponding angles. 
It will be seen from the table that this is not the case with 
the logarithmic sines, tangents, and cotangents of angles 
near to 0°, nor with the logarithmic cosines, tangents, and 
cotangents of angles near to 90°. 

Thus, the difference for 1' in the case of the logarithmic 
sine or tangent of an angle between 40' and 50' is 96.9, 
while for an angle between 50' and 1° it is 79.2. 

A very accurate method for finding the logarithmic sine or 
tangent of an angle near to 0°, or the logarithmic cosine or 
cotangent of an angle near to 90°, is to first calculate the 
natural function by aid of the table of natural sines and 
cosines, or of natural tangents and cotangents, and then 
find the logarithm of the result. 

To find the angle corresponding in similar cases, find the 
number corresponding to the logarithmic function, and then, 
by aid of the tables of natural functions, calculate the angle 
corresponding to the result. 

Ex. 1. Eind log sin 0° m\ 

From the table of natural sines and cosines, we obtain 

natural sin 0° 56' = .016289. 

Whence, log sin 0° 56' = 8.2119 - 10. 

This result is correct to the last place of decimals ; by the ordinary 
method we should have obtained 8.2102 — 10. 

Ex. 2. Eind the angle whose log tan = 8.0302 - 10. 

The number corresponding to this logarithm is .01072. 
From the table of natural tangents and cotangents, the angle whose 
natural tangent is .01072 is 36.85'. 



lo Use of the Tables. 

This is correct to the last place of decimals ; the ordinary method 
would have given 37.15'. 

Note XII. To find with accuracy the log cotangent of an angle 
near to 0°, find the log tangent of the angle by the above method, and 
then take the cologaritlim of the result. (See Note X., page 7.) 

To find the angle corresponding to a log cotangent in a similar case, 
find the log tangent of the angle (Note X.) , and then find the angle 
corresponding as above. 

Note XIII. To find the log tangent of an angle near to 90°, find 
the log tangent of its complement^ and take the cologarithm of the 
result. (See Note XIL) 

To find the angle corresponding in a similar case, find the angle 
corresponding to its cologarithm, and take the complement of the 
result. 

Note XIV. The more accurate method should be employed in 
finding the log sines, tangents, or cotangents of angles between 0° and 
5°, or the log cosines, tangents, or cotangents of angles between 85° 
and 90°, and the angles corresponding in similar cases. For angles 
between 5° and 85° the ordinary method is sufficiently exact. 



FOUR PLACE LOGARITHMIC TABLES 



12 



LOGARITHMS OF NUMBERS. 



No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


lO 




0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


II 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1 106 


13 


"39 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


14 


1 461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


41 16 


4133 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


501 1 


5024 


5038 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 




5302 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


49 


6902 


691 1 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


54 ! 7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



LOGARITHIMS OF NUMBERS. 



13 



No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7S96 


7903 


7910 


7917 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


65 


8l2Q 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


68 


8325 


8331 


8338 


834^ 


8351 


8357 


8363 


8370 


8376 


8382 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 


8513 


8519 


8525 


8531- 


8537 


8543 


8549 


8555 


8561 


8567 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


^733 


8739 


8745 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


97 


9868 


9872 


9877 


988r 


9886 


9890 


9894 


9899 


9903 


9908 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948, 


9952 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



14 



LOGARITHMIC SINES, COSINES, 



Angle. 


Sin. 


D.l'. 


Cos. 


D. 1'. 


Tan. 


D. 1'. 


Cot. 




0" 0' 


— CX) 




10.0000 


.0 


— 00 


301. 1 

176.I 
124.9 
96.9 
79.2 
67.0 
58.0 
51.2 

45-7 
41^5 
37-8 
34^8 
32.2 
30.0 
28 I 


00 


90° 0' 

89° 50' 


74637 


.0000 


7-4637 


2-5363 


0° 20' 

o° 30' 


.7648 
.9408 


176.0 


.0000 
.0000 


.0 


.7648 
.9409 


•2352 
•0591 


89° 40' 
89° 30' 


0" 40' 


8.0658 


96.9 
79.2 
66.9 
58.0 

45.8 
41-3 
37^8 
34-8 
32.1 
30.0 
28 


.0000 


.0 


8.0658 


1.9342 


89° 20' 


0° 50' 


.1627 


.0000 


.1627 


•8373 


89° 10^ 


1° 0' 

1° 10' 


8.2419 


9.9999 
•9999 


.0 


8.2419 


I.7581 


89° 0' 

88° so' 


.3088 


.3089 


.6911 


1° 20' 


.3668 


•9999 




.3669 


•6331 


88° 40' 


1° 30' 


.4179 


.9999 


.1 
.0 
.1 


.4181 


.5819 


88° 30/ 


1° 40' 


•4637 


•9998 


.4638 


.5362 


88° 20' 


1° 50' 


•5050 


■9998 


•5053 


•4947 


88° 10' 


2° 0' 


8.5428 


9.9997 





8.5431 


1.4569 


88° 0' 


2° 10' 


•5776 


•9997 


.1 


•5779 


,4221 


^f. 5°; 


2° 20' 


.6097 


.9996 


.6101 


•3899 


87° 40' 


2° 30' 


•6397 


.9996 




.6401 


•359a 


87° 30' 


2° 40' 


.6677 


26.3 
24.8 

23^5 
22.2 

-7 T -7 


•9995 





.6682 


26.3 
24.9 

23-5 
22.3 
21.3 


•3318 


87° 20' 


2° 50' 
3° 0' 

3° 10' 


.6940 


•9995 


.1 
.1 

.0 

T 


.6945 


•3055 


87° 10' 

87° 0' 

86° 50' 


8.7188 
.7423 


9.9994 
•9993 


8.7194 
•7429 


1.2806 


•2571 


3° id 


•7645 


•9993 


.7652 


.2348 


86° 40' 


3° 30; 


■7857 




•9992 




.7865 


•2135 


86° 30' 


3° 40' 


.8059 


19.2 
18.5 
17.7 
17.0 
16.3 
15.8 


.9991 




.8067 


19.4 
18.5 
17.8 

17.1 
16.5 
15.8 


•1933 


86° 20' 


3050' 
40 0' 
4° 10' 


.8251 
8.8436 


.9990 
9.9989 


.1 
.0 


.8261 


•1739 


86° 10' 
86° 0' 

85° 50; 


8.8446 
.8624 


I-1554 


■8613 


.9989 


•1376 


4° 20' 


•8783 


.9988 


•^ 


•8795 


.1205 


85° 40' 


40 30' 


.8946 


.9987 


■ 


.8960 


.1040 


85° 30; 


40 40' 


.9104 


.9986 




.9118 


.0882 


85° 20' 


4" SO' 
5° 0' 
5° lo' 


.9256 

8.9403 

•9545 


15.2 
14.7 
14.2 
137 


.9985 
9.9983 


.2 
.1 
.1 


.9272 

8.9420 

•9563 


15-4 
14.8 

14-3 

13.8 

13-5 
13.0 
12.7 
12.3 
12.0 
II. 7 
11.4 
III 


.0728 
1.0580 


85° 10' 
85° 0' 

84° 50' 


.9982 


•0437 


5° 2d 


.9682 


.9981 


.9701 


.0299 


84° 40' 


50 30' 


.9816 


13-4 
12.9 
12.5 

12.2 
II.9 

ii-S 
II-3 
10.9 
10.7 
10.4 
10.2 


.9980 


T 


•9836 


.0164 


84° 30' 


50 40' 
5050' 

6° 0' 

6° 10' 


•9945 
9.0070 


•9979 
•9977 


.2 
.1 
.1 


.9966 
9.0093 
9.0216 

.0336 


.0034 
0.9907 
0.9784 

.9664 


84° 20' 
84° lo' 

84° 0' 
83° 50' 


9.0192 


9.9976 
•9975 


.0311 


6° 20' 


.0426 


•9973, 


I 


•0453 


-9547 


83° 40' 


6° 30' 


•0539 


.9972 


T 


.0567 


•9433 


83° 30' 


6° 40' 


.0648 


.9971 





.0678 


10 8 


•9322 


83° 20/ 


6° 50' 


•0755 


.9969 




.0786 


10.5 
10.4 


.9214 


83° 10' 


70 0/ 


9.0859 


9.9968 




9.0891 


0.9109 


83° 0' 


70 10' 


.0961 


9.9 


.9966 


.2 


•0995 


.9005 


82° 50' 


7° 20' 


.1060 


.9964 


.1096 


9.8 


.8904 


82° 40' 


7° 30' 


•1157 


9-7 


•9963 




.1194 


.8806 


82° 30' 




Cos. 


D.l'. 


Sin. 


D. 1'. 


Cot. 


i>. 1'. 


Tan. 


Angle. 



TANGENTS AND COTANGENTS. 



15 



Angle. 


Sin. 


D.l'. 


Cos. 


D. 1'. 


Tan. 


D. 1'. 


Cot. 




7° 30' 


9-1157 




9.9963 


.2 


9.II94 


9-7 
9.4 

9-3 

9-1 

8.9 

^■1 
8 6 


0.8806 


82° 30' 


7° 40' 


.1252 


9-5 
9-3 
9-1 
8.9 
8.7 

8.4 
8.2 
8.0 

7-9 

7-6 
7-5 
1-Z 
7-3 
7-1 
7.0 
6 8 


.9961 


.1291 


.8709 


82° 20' 


7° 50' 
8° 0' 


•1345 
9.1436 


•9959 


.1 

.2 


-1385 


.8615 


82° 10^ 
82° 0' 


9-9958 


9.1478 


0.8522 


8° 10' 


.1525 


.9956 




.1569 


-8431 


81° 50/ 


8° 20' 


,1612 


-9954 





.16S8 


-8342 


81° 40^ 


8° 30' 


.1697 


.9952 


2 


•1745 


•8255 


81° 30' 


8° 40' 


.1781 


-9950 




-I831 


8.4 
8.2 

8.1 


.8169 


81° 20' 


8° 50' 
9° 0' 


.1863 


.9948 
9.9946 


.2 
.2 


•I915 


.8085 


81° 10' 
81° 0' 


9-1943 


9.1997 


0.8003 


9° 10' 


.2022 


-9944 




.2078 


80 


.7922 


80° 50' 


9° 20' 


.2100 


.9942 


.2 



.2158 


7-8 
7-7 
7.6 

7-4 

7-3 
7^1 
7.0 

6.9 
6.8 

6.6 


.7842 


80° 40' 


9° 30' 


.2176 


.9940 


.2236 


.7764 


80° 30/ 


9° 40' 


.2251 


-9938 


.2 
.2 

•3 




•2313 


•7687 


80° 20' 


9° 50' 
10° 0' 


.2324 


.9936 


.2389 


.7611 


80° 10' 
80° 0' 


9-2397 


9-9934 


9.2463 


0-7537 


10° 10' 


.2468 


-9931 


.2536 


.7464 


79° 50' 


10° 20' 


•2538 


.9929 


2 


.2609 


•7391 


79° 40' 


10° 30' 


.2606 


68 


.9927 


•3 
.2 

•3 
.2 


.2680 


•7320 


79° 30' 


10° 40' 
10° 50' 

11° 0' 


.2674 
.2740 


6.6 
6.6 

6.4 

6.4 

6.3 
6 I 


-9924 
.9922 


.2750 
.2819 


.7250 
.7181 


79° 20' 
79° 10' 
79° 0' 


9.2806 


9.9919 


9.2887 


0.7113 


11° 10' 


.2870 


.9917 


•3 


•2953 


6.7 
6-5 
6.4 
6-3 
6-3 
6.1 
6 I 


.7047 


78° 50^ 


11° 2d 


•2934 


.9914 


.3020 


.6980 


78° 40' 


11° 30' 


.2997 


.9912 


•3 
2 


.3085 


.6915 


78° 30' 


11° 40' 


•3058 


6 I 


.9909 


-3149 


.6851 


78° 20' 


ii°5cy 


-3119 


60 


.9907 


•3 

•3 
2 


.3212 


.6788 


78° 10' 


12° 0' 

12° icy 


9-3179 
-3238 


5^9 
5-8 
5-7 

X!^ 

5-5 
5-4 
5-4 
5-3 
5^2 
5-2 
5-1 
5-0 

5-0 
4-9 
4-9 
4.8 

4-7 


9.9904 


9-3275 


0.6725 


78° 0' 

77° 50' 


.9901 


-3336 


.6664 


12° 20' 


.3296 


-9899 


•3 
•3 
•3 
•3 
-3 
•3 
•3 
•3 
.3 
•3 
•3 
•3 
-4 
-3 
•3 
•4 


-3397 


6 I 


.6603 


77° 40' 


12° 30' 


•3353 


.9896 


-3458 


5-9 
5-9 
5.8 

5^7 

5^7 
5.6 

5-5 
5-5 
5-4 

5-3 
5-3 
5^3 
5^1 
5^2 
5-1 


.6542 


77° 30' 


12° 40' 


.3410 


-9893 


-3517 


.6483 


77° 20' 


12° 50' 

13° 0' 

13° 10' 


.3466 


.9890 


-3576 


.6424 


77° lO^ 
77° 0' 
76° 50' 


9-3521 
•3575 


9.9887 


9-3634 
.3691 


0.6366 


.9884 


.6309 


13° 2cy 


.3629 


.9881 


•3748 


.6252 


76° 40' 


130 30, 


.3682 


.9878 


.3804 


.6196 


76° 30' 


13° 40' 


•3734 


-9875 


-3859 


.6141 


76° 20' 


13° 50' 

14° 0' 

14° 10' 


•3786 


.9872 


.3914 


.6086 
0.6032 


76° 10' 
76° 0' 

75° 50' 


9-3837 


9.9869 


9.3968 


-3887 


.9866 


.4021 


•5979 


14° 20; 


-3937 


.9863 


.4074 


.5926 


75° 40' 


14° 30' 


..3986 


•9859 


.4127 


•5873 


75° 30' 


14° 40' 


•4035 


.9856 


.4178 


.5822 


75° 20^ 


14° 50' 
15° 0' 


.4083 


-9853 


.4230 


•5770 


75° 10' 
75° 0' 


9.4130 


9-9849 


9.4281 


0.5719 




Cos. 


D.l'. 


Sin. 


D.l'. 


Cot. 


D. V. 


Tan. 


Angle. 



16 



LOGARITHMIC SINES, COSINES, 



Angle. 


Sin. 


D.l'. 


Cos. 


D. V. 


Tan. 


D. 1'. 


Cot. 




15° 0' 

15° lo' 


9.4130 


4^7 
4.6 
4.6 
4^5 
4-5 
4.4 

4.4 


9.9849 


•3 
■3 


9.4281 


5-0 

5-0 
4.9 
4.9 
4.8 
4.8 

4^7 
4-7 


0-57^9 


75° 0' 

74° 50^ 


•4177 


.9846 


•4331 


•5669 


15° 20^ 


.4223 


•9843 


.4381 


.5619 


74° 40' 


15° 30; 


.4269 


•9839 


•4 
•3 

•4 
•4 
•3 


•4430 


•5570 


74° 30' 


15° 40 


•4314 


.9836 


•4479 


•5521 


74° 20' 


15° 50' 
16° 0' 


•4359 
9.4403 


.9832 


•4527 


•5473 


74° 10' 

74° 0' 


9.9828 


9^4575 


0-5425 


16° 10' 


•4447 


.9825 


.4622 


•5378 


73° 50' 


16° 20' 


.4491 


4.4 


.9821 


•4 


.4669 


•5331 


73° 40' 


16° 30' 


•4533 


4.2 

4^3 
4.2 
4.1 
4.1 
4.1 
4.0 
4.0 
4.0 
3^9 
3^9 
3^8 
3.8 
3^7 
3^8 
3-6 

3-7 

ii 

3-5 
3-4 
3-4 
3-4 
3^4 
3-3 
3-3 
3-3 


•9817 


•4 
•3 

•4 

•4* 

•4 

.4 

•4 

•4 


.4716 


4-7 
4.6 
4.6 

4^5 
4-5 
4-5 


.5284 


73° 30^ 


16° 40' 


•4576 


.9814 


.4762 


.5238 


73° 20' 


16° 50' 
17° ■ 0' 

17° 10' 


.4618 
9.4659 


.9810 


.4808 


.5192 


73° 10' 
73° 0' 

72° 50' 


9.9806 
.9802 


9-4853 


0.5147 


.4700 


.4898 


.5102 


17° 20' 


.4741 


•9798 


•4943 


•5057 


72° 40' 


170 30' 


.4781 


•9794 


•4987 


4.4 


•5013 


72° 30' 


17° 40' 


.4821 


.9790 


•5031 


4.4 


•4969 


72° 20' 


17° 50' 
18° 0' 

18° 10' 


.4861 
9.4900 


.9786 


•4 
•4 
•4 
•4 
•4 
•5 


•5075 


4.4 
4^3 
4-3 
4.2 
4.2 
4.2 
4.2 
4^1 
4.1 
4.0 
4.0 
4.0 
4.0 
4.0 

3^9 

3-S 

3.8 
3^8 

3-7 
3^8 
3^7 
3^7 
3^7 
3-6 

3-6 
3-6 
3-6 


•4925 


72° 10' 

72° 0' 
71° 50^ 


9.9782 


9.5118 


0.4882 


•4939 


.9778 


.5161 


.4839 


18° 20' 


•4977 


•9774 


•5203 


•4797 


71° 40' 


18° 30' 


•5015 


.9770 


•5245 


•4755 


71° 30' 


18° 40' 


•5052 


•9765 


.5287 


•4713 


71° 20' 


18° 5(y 


.5090 


.9761 


•4 
,4 

•5 
•4 
•5 
.4 
'5 
•4 

•5 
•4 
•5 
•5 
•5 
•4 

•5 
•5 
•5 

•5 
•5 
•5 

•5 
6 


•5329 


.4671 


71° 10' 


19° 0' 

19° ic/ 


9.5126 
•5163 


9-9757 
•9752 


9^5370 


0.4630 


71° 0' 
70° 50' 


•54" 


•4589 


19° 20' 


•5199 


•9748 


•5451 


•4549 


70° 40' 


19° 30; 


•5235 


•9743 


•5491 


•4509 


70° 30' 


19° 40' 


.5270 


•9739 


•5531 


•4469 


70° 20' 


19° 50' 


•5306 


•9734 


•5571 


.4429 


70° lO^ 


20° 0' 

20° 10' 


9^5341 
•5375 


9-9730 


9.5611 
.5650 


0.4389 


70° 0' 

69° 50' 


•9725 


•435^ 


20° 20' 


•5409 


.9721 


.5689 


•431 1 


69° 40' 


20° 30' 


•5443 


.9716 


•5727 


•4273 


69° 30' 


20° 40' 


•5477 


.9711 


.5766 


.4234 


69° 20' 


20° 50' 
21° 

21° 10' 


•5510 


.9706 


•5804 
9.5842 


.4196 


69° 10' 
69° 0' 

68° 50' 


9^5543 
•5576 


9.9702 


0.4158 


.9697 


•5879 


4121 


21° 20' 


.5609 


3-3 

3-2 
3^2 

3^1 

3^2 

3^1 
3-1 


.9692 


•5917 


.4083 


68° 40' 


21° 30' 


•S641 


.9687 


•5954 


.4046 


68° 30' 


21° 40' 


•5673 


.9682 


•5991 


.4009 


68° 20' 


21° 50' 
22° 0' 


•5704 


.9677 
9.9672 


.6028 


•3972 


68° 10' 
68° 0' 


9^5736 


9.6064 


0.3936 


22" 10' 


•S767 


.9667 


.6100 


.3900 


67° 50' 


22° 20' 


•5798 


.9661 


•5 


.6136 


.3864 


67° 40' 


22° 30' 


.5828 


3-0 


.9656 


.6172 


.3828 


67° 30' 




Cos. 


D.l'. 


Sin. 


D. 1'. 


Cot. 


T>.V. 


Tan. 


Angle. 



TANGENTS AND COTANGENTS. 



17 



Angle. 


Si*». 


D.l'. 


Cos. D 


1'. 


Tan. 


D. \t. 


Cot. 




22° 30' 


9.5828 


3-1 


9.9656 


5 

I 

5 


9.6172 


3-6 
3-5 
3-6 

3-5 
' 3-4 
3-5 
3-4 
3-5 
3-4 

3-4 

ZA 
Z-Z 
3-4 
l-Z 
Z-Z 
3-2 

3^2 

Z-Z 
Z-2 

3^2 
3^2 

3-1 
3-2 
3-1 
3^2 

3^1 
3^1 
3^1 

3^1 
3-0 
3-1 
3-0 
3-0 


0.3828 


67° 30' 


22° 40' 


•5859 


•9651 


.6208 


•3792 


67° 20' 


22° 50' 

23° 0' 

23° 10' 


.5889 


3-0 
3-0 

2.9 

3-0 


.9646 


.6243 


•3757 


67° 10^ 
67° 0' 

66° 50' 


9-5919 


9.9640 
•9635 


9.6279 


0.3721 


•5948 


.6314 


.3686 


23° 20' 


.5978 


.9629 


\ 


.6348 


.3652 


66° 40' 


23° 30' 


.6007 


2.9 
2.9 


.9624 


•6383 


-3617 


66° 30' 


23° 40' 


.6036 


.9618 


5 
6 


.6417 


•3583 


66° 20' 


23° 50' 


.6065 


2.9 
2.8 


.9613 


.6452 


-3548 


66° 10' 


24° 0' 


9.6093 


2.8 


9.9607 


5 
6 


9.6486 


0.3514 


66° 0' 


24° 10' 


.6121 


2.8 
2 8 


.9602 


,6520 


.3480 


65° 50' 


24° 20' 


.6149 


.9596 


6 


•6553 


•3447 


65° 40' 


24° 30' 


.6177 


2.8 
2.7 
2.7 
2.7 
2.7 
2.7 
9 6 


•9590 


6 

5 
6 


.6587 


•3413 


65° 30' 


24° 40' 


.6205 


•9584 


.6620 


•3380 


65° 20' 


24° 50' 


.6232 


•9579 


.6654 


•3346 


65° 10' 


25° 0' 


9.6259 


9-9573 


6 


9.6687 


0^3313 


65° 0' 


25° 10' 


.6286 


-9567 


6 


.6720 


.3280 


64° 50' 


25° 20; 


•6313 


.9561 


6 


.6752 


.3248 


64° 40' 


25° 30' 


.6340 


-9555 


(=1 


•678s 


•3215 


64° 30' 


25° 40' 


.6366 


2.6 

26 


-9549 


6 


.6817 


•3183 


64° 20' 


25° 50' 


.6392 


•9543 


6 


.6850 


•3150 


64° 10' 


26° 0' 


9.6418 


2.6 


9-9537 


7 


9.6882 


0.31 18 


64° 0' 


26° 10' 


.6444 


^ 6 


•9530 


.6914 


.3086 


63° 50' 


26° 20' 


.6470 


2.5 

2 6 


•9524 


6 

6 


.6946 


•3054 


63° 40' 


26° 30' 


•6495 


.9518 


.6977 


•3023 


63° 30' 


26° 40' 


.6521 


2-5 

2.4 
2.5 

2-5 


•9512 


7 
6 


.7009 


.2991 


63° 20' 


26° 50' 


.6546 


•9505 


.7040 


.2960 


63° 10' 


27° 0' 


9.6570 


9-9499 


7 
6 


9.7072 


0.2928 


63° 0' 


27° icy 


•6595 


.9492 


-7103 


•2897 


62° 50' 


27° 20' 


.6620 


.9486 


7 


•7134 


.2866 


62° 40' 


27° 30' 


.6644 


2.4 

2.4 


•9479 


-7165 


•2835 


62° 30' 


27° 40' 


.6668 


•9473 


7 
7 
6 


.7196 


.2804 


62° 20' 


27° 50' 

28° 0' 


.6692 
9.6716 


2,4 

2.4 
2.4 

2-3 


.9466 


.7226 


.2774 


62° 10' 
62° 0' 


9^9459 


9.7257 


0.2743 


28° 10' 


.6740 


•9453 


7 
7 
7 


.7287 


•2713 


61° 50' 


28° 20' 


.6763 


.9446 


•7317 


.2683 


61° 40' 


28° 30' 


.6787 


2.4 

2-3 


.9439 


•7348 


Z-^ 

3-0 


.2652 


61° 30' 


28° 40' 


.6810 


•9432 


•7378 


.2622 


61° 20' 


28° 50' 


.6833 


2-3 
2.2 


-9425 


7 
7 
7 
7 
7 
7 
7 
8 


.7408 


3-0 
3-0 
2.9 

3-0 

2.9 

3-0 
2.9 
2.9 


•2592 


61° 10' 


29° 0' 


9.6856 


9.9418 


9^7438 


0.2562 


61 °0' 


29° 10' 


.6878 




.9411 


.7467 


•2533 


60° 50' 


29° 2.0' 


.6901 


.9404 


•7497 


•2503 


60° 40' 


29° 30' 


.6923 


^3 


•9397 


.7526 


•2474 


60° 30' 


29° 40' 


.6946 


•9390 


•7S56 


.2444 


60° 20' 


29° 50' 
30° 0' 


.6968 


2.2 


•9383 


•7585 


.2415 


60° 10/ 
60° 0' 


9.6990 


9-9375 


9.7614 


0.2386 




Cos. 


D.l'. 


Sin. D. 


1'. 


Cot. 


D.l'. 


Tan. 


Angle. 



18 



LOGARITHMIC SINES, COSINES, 



Angle. 


Sin. 


D.l'. 


Cos. 


B. 1'. 


Tan. 


D. 1'. 


Cot. 




30° 0' 


9.6990 


2 2 


9-9375 


•7 

-7 
.8 


9.7614 


3-0 
2.9 

2.8 


0.2386 


60° 0' 


30° 10' 


.7012 




.9368 


.7644 


.2356 


59° 50' 


30° 20' 


.7033 




.9361 


•7673 


•2327 


59° 40' 


30° 30; 


•7055 




.9353 


i 


.7701 


2.9 

2-9 

2.9 

2.8 

2.9 

2.8 

2.9 
28 


.2299 


59° 30' 


30° 40' 


7076 


2 I 


•9346 


•7730 


.2270 


59° 20' 


30° 50' 


.7097 


2.1 


-9338 


•7 

.8 

8 


-7759 


.2241 


59° 10' 


31° 0' 

31° 10' 


9.7118 
7139 


2.1 
2 I 


9-9331 


9-7788 


0.2212 


59° O' 

58° 50' 


•9323 


.7816 


.2184 


31° 20' 


.7160 




-9315 




.784s 


•2155 


58° 40' 


31° 30' 


.7181 


2 


.9308 


•7873 


.2127 


58° 30' 


31° 40' 


.7201 




.9300 


8 


.7902 


.2098 


58° 20' 


31° 50' 


.7222 


2 


.9292 


.8 


.7930 


28 


.2070 


58° 10' 


32°' 0' 

32° 10' 


9.7242 
.7262 


2.0 
2.0 


9.9284 
.9276 


.8 


9-7958 
.7986 


2.8 
2.8 
2.8 

28 


0.2042 


58° 0' 

57° 50' 


.2014 


32° 20' 


.7282 


.9268 


S 


.8014 


.1986 


57° 40' 


32° 30' 


.7302 


2 


.9260 


8 


.8042 


.1958 


57° 30' 


32° 40; 


.7322 




.9252 


8 


.8070 


1:1 

2.8 


.1930 


57° 20; 


32° 50' 
33° 0' 


•7342 


1^9 
1.9 


.9244 
9.9236 


.8 
.8 


.8097 


.1903 


57° 10' 

57° 0' 


9-7361 


9.8125 


0.1875 


330 10' 


■7380 


.9228 


•9 
8 


.8153 


2.7 

2 8 


.1847 


56° 50^ 


33° 20' 


.7400 


1.9 
1.9 

1-9 
1.9 

I 8 


.9219 


.8180 


.1820 


56° 40' 


33° 30' 


.7419 


.9211 


.8 

-9 
.8 


.8208 


^1 


.1792 


56° 30' 


33° 40' 


•7438 


.9203 


.8235 


■1765 


56° 20' 


33° so' 


•7457 


.9194 


.8263 


2.7 
2.7 

2-7 
2.7 

2-7 

2-7 
2.7 

2.7 

2.7 


•1737 


56° 10' 


34° 0' 


9.7476 


9.9186 


•9 
.8 
•9 
-9 
•9 
.8 

•9 
•9 
•9 
•9 
•9 
•9 

I.O 

.9 
-9 

T n 


9.8290 


0.I7I0 


56° 0' 


34° icy 


•7494 


1.9 

1.8 


.9177 


-8317 


.1683 


55° 50' 


34° 20' 


•7513 


.9169 


-8344 


.1656 


55° 40' 


34° 30' 


•7531 


.9160 


-8371 


.1629 


55° 30' 


34° 40' 


•7550 


1.9 

I 8 


-9151 


.8398 


.1602 


55° 20' 


34° 50' 
35° 0' 

3S° 10' 


.7568 


1.8 
1.8 
1.8 
1.8 
1.7 
1.8 
1-7 
1.8 

1-7 
1-7 
1-7 
1-7 
1-7 
1.6 

1-7 

1.6 


.9142 


-8425 
9-8452 


•1575 


55° 10/ 
55° 0' 

54° 50' 


9.7586 


9-9134 
.9125 


0.1548 


.7604 


•8479 


.1521 


35° 20' 


.7622 


.9116 


.8506 


.1494 


54° 40' 


35° 30' 


.7640 


.9107 


.8533 


.1467 


54° 30' 


35° 40' 


•7657 


.9098 


-8559 


2.7 

2-7 
2.6 

"I 


.1441 


54° 20' 


35° 50' 


•7675 


.9089 


.8586 


.1414 


54° 10' 


36° 0' 

36° 10' 


9.7692 


9.9080 
.9070 


9.8613 


0.1387 


54° 0' 
53° 50' 


.7710 


.8639 


.1361 


36° 20' 


.7727 


.9061 


.8666 


•1334 


53° 40' 


36° 30' 


■7744 


.9052 


.8692 


2 6 


.1308 


53° 30' 


36° 40' 


.7761 


.9042 


•9 
1.0 

•9 


.8718 


2-7 
2.6 

2.6 

2-7 
2.6 


.1282 


53° 20' 


36° 50^ 
37° 0' 
37^ 10' 


.7778 


-9033 


-8745 


•1255 


53° 10' 
53° 0' 

52° 50' 


9-7795 
.7811 


9.9023 


9.8771 


0.1229 


.9014 


.8797 


.1203 


37° 20; 


.7828 


.9004 


•9 


.8824 


.1176 


52° 40' 


37° 3o'' 


.7844 


•8995 


.8850 


.1150 


52° 30' 




Cos. 


D.l'. 


Sin. 


D.l'. 


Cot. 


D. 1'. 


Tan. 


Angle. 



TANGENTS AND COTANGENTS.. 



19 



Angle. 


Sin. 


D.l'. 


Cos. 


D.l'. 


Tan. 1 


D.l'. 


Cot. 




370 30' 


9.7844 


"A 


9.8995 




9.8850 


2.6 
2.6 
2.6 
2.6 
2 6 


0.II50 


52° 30' 


37° 40' 


.7861 


.8985 




.8876 


.1124 


52° 20' 


37° 50' 
38° 0' 

38° 10' 


.7877 


1.6 

1^7 

I 6 


.8975 
9.8965 


I.O 
I.O 


.8902 


.1098 


52° 10' 

52° 0' 

51° 50^ 


9-7893 


9.8928 


0.1072 


.7910 


-8955 


.8954 


.1046 


38° id 


.7926 


1^5 
I 6 


.8945 




.8980 


2 6 


.1020 


51° 40' 


38° 30' 


.7941 


.8935 




.9006 


2 6 


.0994 


51° 30' 


38° 40' 


•7957 


I 6 


.8925 




.9032 


26 


.0968 


51° 20' 


38° 50' 
39° 0' 


•7973 


1.6 

1.6 
1-5 
1-5 

T 6 


.8915 
9.8905 


1.0 

I 


.9058 


2.6 

26 


.0942 


51° 10' 

51° 0' 


9.7989 


9.9084 


0.0916 


39° 10' 


.8004 


.8895 




.9110 


2 6 


.0890 


50° 50' 


39° 20' 


.8020 


.8884 




•9135 


.0865 


50° 40' 


39° 30' 


•8035 


-8874 




.9161 


.0839 


50° 30' 


39° 40' 


.8050 


.8864 




.9187 


ri 


.0813 


50° 20' 


39° 50' 


.8066 


1-5 
1-5 
1-5 
1^4 
15 
1^5 
1.4 

1-5 
1.4 

1-5 
1.4 
1.4 
1.4 
1.4 
1.4 
1.4 
1.4 

1-3 
1.4 

1-3 
1-4 
1-3 
1-3 
1-4 
1-3 
1-3 
1-3 
1-3 


•8853 


I.O 


.9212 


.0788 


50° 10' 


40° 0' 


9.8081 


9-8843 




9.9238 


26 


0.0762 


50° 0^ 


40° 10' 


.8096 


.8832 




.9264 


"1 


.0736 


49° 50' 


40° 20' 


.8111 


.8821 




.9289 


.0711 


49° 40' 


40° 30' 


•8125 


.8810 


T 


-9315 


ft 


.0685 


49° 30' 


40° 40' 


.8140 


.8800 


I.I 
I.I 
I.I 


-9341 


2-5 
2.6 

2^5 
2 6 


.0659 


49° 20' 


40° 50' 
41° 0' 
41° 10' 


•8155 


.8789 


.9366 


.0634 


49° 10' 
49° 0' 

48° 50/ 


9.8169 


9.8778 


9.9392 


0.0608 


.8184 


.8767 


.9417 


.0583 


41° 20' 


.8198 


-87S6 




-9443 


2^5 
2.6 


•0557 


48° 40' 


410 30; 


•8213 


•8745 




.9468 


.0532 


48° 30' 


41° 40' 


.8227 


-8733 




-9494 


.0506 


48° 20' 


41° 50^ 
42° 0' 
42° \d 


.8241 


.8722 

9.8711 

.8699 


I.I 

1.2 
J J 


•9519 


2^5 
2-5 
2.6 

2-5 
2.6 

2-5 


.0481 

0.0456 


48° 10' 
48° 0' 

47° 50' 


9-8255 


9-9544 


.8269 


-9570 


.0430 


42° 20; 


.8283 


.8688 




•9595 


.0405 


47° 40' 


42° 30 


.8297 


.8676 


T T 


.9621 


•0379 


47° 30' 


42° 40' 


.8311 


.8665 




.9646 


•0354 


47° 20' 


42° 50' 
43° 0' 


.8324 


.8653 


1.2 
I 2 


.9671 


2-5 
2.6 

2-5 
2^5 


.0329 

0.0303 


47° 10' 

47° 0' 


9^8338 


9.8641 


9.9697 


43° 10; 


•8351 


.8629 


T T 


.9722 


.0278 


46° 50' 


430 20 


•8365 


.8618 


I 2 


•9747 


•0253 


46° 40' 


43° 30' 


.8378 


.8606 




.9772 


.0228 


46° 30' 


43° 40; 


.8391 


-8594 


1.2 
1-3 
1.2 
1.2 


.9798 


.0202 


46° 20' 


43° 50' 
44° 0' 
44° 10' 


.8405 


.8582 


.9823 


2^5 
2-5 
2.6 


.0177 


46° 10' 
46° 0' 

45° 50' 


9.8418 
.8431 


9.8569 
-8557 


9.9848 


0.0152 


•9874 


.0126 


44° 20' 


.8444 


•8545 


•9899 


2-5 


.0101 


45° 40' 


44° 30' 


•8457 


•8532 


^^•3 


.9924 


2^5 

2.5 

2.6 

2-5 


.0076 


45° 30' 


44° 40' 


.8469 


1-3 
1-3 


.8520 




•9949 


.0051 


45° 20' 


44° 50' 
45° 0' 


.8482 


.8507 


1-3 
1.2 


-9975 
0.0000 


.0025 


45° 10' 

45° 0' 


9.8495 


9-8495 


0.0000 




Cos. 


D.l'. 


Sin. 


D. 1'. 


Cot. 


D. 1'. 


Tan. 


Angle. 



20 



NATURAL SINES AND COSINES. 



A. 


Sin. 


Cos. 




A. 


Sin. 


Cos. 




A. 


Sin. 


Cos. 




0° 
lo' 
20' 
30; 
40' 
50' 
1° 
10' 
20' 
30; 
40' 

50' 
2° 
10' 
20' 
30' 
40' 
50' 
3° 
10' 
20' 

30' 
40; 

50' 

40 

10' 
20' 
30' 
40' 
50' 
5° 
10' 
20' 
30; 
40' 
50' 
6° 
10' 
20' 
30; 
40' 
50' 
7° 
10' 
20' 
30' 


.000000 I 


0000 


90° 

50; 
40' 
30' 
20' 
10' 

50' 
40' 
30; 
20' 
10' 
88° 
50; 
40' 
30; 
20' 
10' 
87° 
50; 
40' 
30; 
20' 
10' 
86° 
50; 
40' 
30' 
20' 
10' 

85° 

50; 
40' 
30; 
20' 
10' 
84° 
50; 
40' 
30' 
20' 
10' 
83° 
50; 
40' 
30' 


30; 
40' 
50' 
8° 
10' 
20' 
30; 
40' 

50' 
9° 

10' 

20' 
30; 
40' 

50' 
10° 

10' 

20' 

30' 

40; 

50' 

11° 

10' 

20' 

30; 

40' 

50' 
12° 
10' 
20' 

30; 
40 

50^ 
13° 

10' 

20' 

30' 

40' 

50' 

14° 

10' 

20' 

30; 

40' 

50' 
15° 


•1305 
•1334 
•1363 


.9914 
.9911 
•9907 


30' 
20' 
10' 

82° 
50' 
40' 
30; 
20' 
10' 
81° 
50; 
40' 
30; 
20' 
10' 
80° 
50; 
40' 

30' 
20' 
10' 

79° 

50; 

40' 

30' 
20' 
10' 

78° 

50; 

40 

30 

20' 

10' 

77° 

50; 

40' 

30' 

20' 

10' 

76° 

50; 

40' 

30; 

20' 

75° 


15° 

10' 

20/ 

30; 

40' 

50' 

16° 

10' 

20' 

30' 

40' 

50' 

17° 

10' 

20' 

30; 

40' 

50' 

18° 

10' 

20' 

30' 

40' 

50' 

19° 

10' 

20' 

30; 

40' 

50' 

20° 

10' 

20' 

30' 

40' 

50' 

21° 

10' 

20' 

30; 

40' 

50' 
22° 
10' 
20' 
30' 


.2588 


•9659 


75° 
50; 
40' 

30' 
20' 
10' 
74° 
50; 
40' 
30' 
20' 
10' 
73° 

sc 

40' 

30; 

20' 

10' 

72° 

50' 

40' 

30' 

20' 

10' 

71° 

50; 
40' 
30; 
20' 
10' 
70° 

^^ 
40^ 

30' 
20' 
10' 
69° 

50' 

40' 

30' 

20' 

10' 

68° 

50; 

40 

30' 

A. 


.002909 I 
.005818 I 
.008727 I 
.011635 
.014544 


0000 
0000 
0000 
9999 
9999 


.2616 
.2644 
.2672 
.2700 
.2728 


•9652 
.9644 
•9636 
.9628 
.9621 


.1392 


•9903 


.1421 

.1449 
.1478 
•1507 

•1536 


.9899 
•9894 
.9890 
.9886 
.9881 


.017452 


9998 


.2756 


.9613 


.02036 
.02327 
.02618 
.02908 
.03199 


9998 
9997 
9997 
9996 
9995 


.2784 
.2812 
.2840 
.2868 
.2896 


.9605 
.9596 
.9588 
.9580 
•9572 


.1564 


.9877 


•1593 
.1622 
.1650 
.1679 
.1708 


.9872 
.9868 
•9863 
.9858 

.9853 


.03490 


9994 


.2924 


•9563 


.03781 
.04071 
.04362 
•04653 
■04943 


9993 
9992 

9990 
9989 
9988 


.2952 
.2979 
•3007 

•3035 
.3062 


•9555 
•9546 
•9537 
•9528 
•9520 


•1736 


.9848 


•1765 
.1794 
.1822 
.1851 
.1880 


•9843 
.9838 
•9833 
•9827 
.9822 


.05234 


9986 


•3090 


•95" 


•05524 
•05814 
.06105 

•06395 
.06685 


9985 
9983 
9981 
9980 
9978 


.3118 
•3145 
•3173 
.3201 
.3228 


•9502 
•9492 
•9483 
•9474 
•9465 


.1908 


.9816 


•1937 
.1965 
.1994 
.2022 
.2051 


.9811 
.9805 
•9799 
•9793 
•9787 


.06976 


9976 


.3256 


•9455 


.07266 

•07556 
.07846 
.08136 
.08426 


9974 
9971 
9969 
9967 
9964 


•3283 
.33^1 
•3338 
•3365 
•3393 


.9446 
•9436 
.9426 
.9417 
•9407 


.2079 


.9781 


.2108 
.2136 
.2164 
.2193 
.2221 


•9775 
.9769 

•9763 
•9757 
.9750 


.08716 


9962 


.3420 


•9397 


.09005 
.09295 

•09585 
.09874 
.10164 


9959 
9957 
9954 
9951 
9948 


•3448 

•3475 
•3502 
•3529 
•3557 


•9387 
•9377 
•9367 
•9356 
■9346 


.2250 


•9744 


.2278 
.2306 
•2334 
■2363 
•2391 


•9737 
•9730 
.9724 
.9717 
.9710 


.10453 


9945 


•3584 


•9336 


.10742 
.11031 
.11320 
. 1 1 609 
.11898 


9942 
9939 
9936 
9932 
9929 


.3611 
.3638 
•3665 
.3692 

•3719 


•9325 
•9315 
•9304 
•9293 
.9283 


.2419 


•9703 


.2447 
.2476 
•2504 

•2532 
.2560 


.9696 
.9689 
.9681 

•9674 
.9667 


.12187 


9925 


•3746 


.9272 


.12476 
.12764 
•13053 


9922 
9918 
.9914 


•3773 
.3800 
.3827 


.9261 
•9250 
•9239 


.2588 


•9659 




Cos. 


Sin. 


A. 




Cos. 


Sin. 


A. 




Cos. 


Sin. 



NATURAL SINES AND COSINES. 



21 



A. 



Sin. 



40' 
50' 

23° 

10' 

20' 

30' 

40' 

50' 

24° 
10' 
20' 
30; 
40' 
50' 
25° 
10' 
20' 
30' 
40' 
50' 
26° 
10' 
20' 
30' 
40' 
50' 

27° 
10' 
20' 
30' 
40' 
50' 
28° 
10' 
20' 
30' 
40' 
50' 

29° 

10' 
20' 
30' 
40' 
50' 
30° 



3827 
3854 
3881 



3907 



3934 
3961 

3987 
4014 
4041 



4067 



4094 
4120 
4147 

4173 
4200 



4226 



4253 
4279 
4305 
4331 
4358 



4384 



4410 
4436 
4462 



4514 



4540 



4566 
4592 
4617 

4643 
4669 



4695 



4720 
4746 
4772 

4797 
4823 



4874 
4899 
4924 
4950 
4975 



5000 



Cos. 



9239 
9228 
9216 



9135 



9205 



9194 
9182 
9171 

9159 
9147 



9124 
9112 
9100 



9075 



9063 



9051 
9038 
9026 
9013 
9001 



8975 
8962 

8949 
8936 
8923 



8910 



8897 



8857 
8843 



8816 
8802 



8774 
8760 



8746 



8732 
8718 
8704 



8675 



8660 



Cos. 



Sin. 



30' 
20' 
10' 
67° 
50; 
40' 
30' 
20' 
10' 
66° 
50' 
40' 
30' 
20' 
10' 
65° 
50' 
40' 
30' 
20' 
10' 
64° 

50' 
40' 
30' 
20' 
10' 
63° 

scy 
40' 
30' 
20' 
10' 

62° 

50' 
40' 
30' 
20' 
10' 
61° 
50' 
40' 
30' 
20' 
10' 
60^ 



A. 



30° 

10' 

20' 

30' 

40' 

50' 

31° 

10' 

20' 

30' 

40' 

50' 

32° 
10' 
20' 
30' 
40' 
50' 

33° 

10' 
20' 
30' 
40' 
50' 
34° 



30' 
40' 
50/ 

35° 

10' 
20' 
30' 
40' 
50' 

36° 

10' 
20' 
30' 
40' 
50' 
37° 
10' 
20' 
30' 



Sin. Cos. 



5000 



5025 
5050 
5075 
5100 

5125 



5150 



5175 
5200 

5225 
5250 
5275 



5299 



5324 
5348 
5373 
5398 
5422 



5446 



5471 
5495 
5519 

5544 
5568 



5592 



5616 
5640 
5664 
5688 
5712 



5736 



5760 
5783 
5807 
5831 
5854 



5878 



5901 
5925 
5948 
5972 
5995 



6018 



6041 
6065 



8660 



8646 
8631 
8616 
8601 

8587 



8572 



8557 
8542 
8526 
8511 
8496 



8480 



8465 
8450 

8434 
8418 
8403 



8387 



8371 
8355 
8339 
8323 
8307 



8290 



8274 
8258 
8241 
8225 
8208 



I192 



8175 
8158 
8141 
8124 
8107 



8090 



8073 
8056 



8021 
8004 



7986 



7969 
7951 
7934 



Cos. Sin. 



60° 

50' 
40' 
30' 
20' 
10' 
59° 

50' 

40' 

30' 

20' 

10' 

58° 

50; 

40' 

30' 

20' 

10' 

57° 
50; 
40' 

30' 
20' 
10' 
56° 

50' 
40' 
30' 
20' 
10' 
55° 
50' 
40' 
30' 
20' 
10' 
54° 
50; 
40' 
30' 
20' 
10' 
53° 
50; 
40' 
30' 



A. 



A. 



3c/ 
40' 

50' 

38° 

10' 

2(y 
30' 
40' 
50' 
39° 



20' 

30' 
40' 
50' 

40° 

10' 
20' 
30' 
40' 
50' 

41° 

10' 
20' 
30' 
40' 
50' 
42° 
10' 
20' 
30' 
40' 
50' 
43° 
10' 
20' 
30' 
40' 
50' 

44° 
10' 
20' 
30' 
40' 
50' 
45° 



Sin. Cos. 



6088 
61II 
6134 



6157 



6180 
6202 
6225 
6248 
6271 



6293 



6316 
6338 
6361 

6383 
6406 



6428 



6450 
6472 
,6494 
6517 
6539 



6561 



,6583 
,6604 
,6626 
,6648 
6670 



669] 



6713 
6734 
,6756 
6777 
6799 



6820 



.6841 
,6862 
,6884 
,6905 
.6926 



6947 



6967 



,7009 
,7030 
,7050 



7071 



7934 
7916 

7898 



7862 
7844 
7826 
7808 
7790 



7771 



7753 
7735 
7716 
7698 
7679 



7660 



7642 
7623 
7604 

7585 
7566 



7547 



7528 

7509 
7490 
7470 

7451 



7431 



7412 
7392 
7373 
7353 
7333 



7314 



7294 
7274 

7254 
7234 
7214 



7193 



7173 
7153 
7133 
7112 
7092 



7071 



30' 
20' 
10' 
52°' 

50' 
40' 
30' 
20' 
10' 
51° 
50; 
40' 
30' 
20' 
10' 
50° 
50' 
40' 
30' 
20' 
10' 
49° 

50' 
40' 



48° 

50; 
40' 
30' 
20' 
10' 

47° 
50; 
40' 
30' 
20' 
10' 
46^ 

50' 
40' 
30' 
20' 
10' 
45^ 



Cos. Sin. 



22 



>^ATURAL TANGENTS AND COTANGENTS. 



A. 


Tan. 


Cot. 




A. 


Tan. 


Cot. ! 


A. 


Tan. 


Cot. 




0° 

lO' 

20' 

30' 
50' 

1° 
10' 
20' 

30; 

40' 
50' 

2° 
10' 
20' 
30' 
40' 
50' 
3° 
10' 
20' 
30' 
40; 
50' 
4° 
10' 
20' 
30; 
40' 
50' 
5° 
10' 
20' 

30; 
40' 
50' 
6° 

10' 
20' 
30; 
40' 
50' 
7° 
10' 
20' 
30' 


.000000 


00 


90° 

50; 
40' 
30^ 
20' 
10' 
89° 
50; 
40' 
30' 
20' 
10' 
88° 
50; 
40' 
30' 
20' 
10' 

87° 

50; 
40' 

30' 
20' 
10' 
86° 

^^' 
10' 

85° 

50; 
40' 
30; 
20' 
10' 
84° 
50; 
40' 
30' 
20' 
10' 
83° 
50; 
40' 
30' 


30' 
40' 
50' 
8° 
10' 
20' 
30; 
40' 

50' 

90 

10' 

20' 

30; 

40^ 

50' 

10° 

10' 

20' 

30' 

40' 

50' 

11° 

10' 

20' 

30; 

40 

50' 

12° 

10' 

20' 

30' 

40; 

50' 

13° 

10; 

20' 

30; 

40' 

50' 

14° 

10' 

20' 

30; 

40' 

50' 

15° 


•I317 
.1346 

-1376 


7^5958 
7.4287 
7.2687 


30; 
20' 
10' 

82° 

50' 

40' 

30' 
20' 
10' 
81° 
50; 
40' 
30' 
20' 
10' 
80° 
50; 
40' 
30' 
20' 
10' 
79° 
50; 
40' 

30; 
20' 
10' 

78° 
50' 
40' 
30' 
20' 
10' 
77° 
50; 
40' 
30' 
20' 
10' 
76° 

50' 
40' 
30; 
20' 
10' 
75° 


15° 

10' 
20^ 
30' 
40' 
50' 
16° 
10' 
20' 
30; 
40' 

50' 
17° 
10' 
20' 
30; 
40' 
50' 
18° 
10' 
20' 
30' 
40' 
50' 
19° 
10' 
20' 
30; 
40' 
50' 
20° 
10' 
20' 
30; 
40' 
50' 
21° 
10' 
20' 
30; 
40' 

50' 
22° 
10' 
20' 
30' 


.2679 


3^732i 


75° 

50' 

40' 

30' 

20' 

10' 

74° 

50; 

40' 

30' 

20' 

10' 

73° 

50; 

40' 

30' 

20' 

10' 

72° 

50; 

40' 


.002909 
.005818 
.008727 
.011636 
.014545 


343-7737 
171.8854 
114.5887 

85.9398 
68.7501 


.2711 

.2742 

-2773 
.2805 
.2836 


3.6891 
3.6470 
3-6059 
3-5656 
3.5261 


.1405 


7-II54 


•1435 
.1465 

■1495 
.1524 

•1554 


6.9682 

6.8269 

6.6912 
6.5606 

6.4348 


■017455 


57.2900 


.2867 


3-4874 


.02036 
.02328 
.02619 
.02910 
.03201 


49.1039 
42.9641 
38.1885 

34-3678 
31.2416 


.2899 
.2931 
.2962 

-2994 
.3026 


3-4495 
3.4124 

3-3759 
3-3402 
3-3052 


.1584 


6.3138 


.1614 
.1644 
■1673 
■1703 
•1733 


6.1970 
6.0844 
5-9758 
5.8708 

5-7694 


.03492 


28.6363 


-3057 


3.2709 


•03783 
.04075 
.04366 
.04658 
.04949 


26.4316 
24.5418 
22.9038 
21.4704 
20.2056 


.3089 
.3121 

•31^3 
-3185 

•3217 


3-2371 
3.2041 
3.1716 

3-1397 
3.1084 


.1763 


5-6713 


■1793 
.1823 

.1914 


5-5764 
5-4845 

5-3955 
5-3093 
5-2257 


.05241 


19.0811 


•3249 


3-0777 


•05533 
.05824 
.06116 
.06408 
.06700 


18.0750 
17.1693 
16.3499 
15.6048 
14.9244 


.3281 
•3314 
•3346 
•3378 
•341 1 


3-0475 
3.0178 
2.9887 
2.9600 
2.9319 


.1944 


5.1446 


30; 1 


.1974 
.2004 

■2035 
.2065 
.2095 


5.0658 
4.9894 
4.9152 
4-8430 
4-7729 


20! 
10' 
71° 

50; 
40' 
30' 
20' 
10' 
70° 

50; 
40' 

30; 
20' 
10' 

69° 

50; 
40 

30 
20' 
10' 
68° 

^°; 
40' 

30' 


.06993 


14.3007 


-3443 


2.9042 


.07285 
.07578 
.07870 
.08163 
.08456 


13.7267 
13.1969 
12.7062 
12.2505 
11.8262 


•3476 
•3508 
•3541 
•3574 
.3607 


2.8770 
2.8502 
2.8239 
2.7980 

2.7725 


.2126 


4.7046 


.2156 
.2186 
.2217 
.2247 
.2278 


4.6382 

4-5736 
4.5107 

4-4494 
4-3897 


.08749 


11.4301 


•3640 


2-7475 


.09042 

•09335 
.09629 
.09923 
.10216 


11.0594 
10.7119 

10.3854 

10.0780 

9.7882 


•3673 
.3706 

•3739 
•3805 


2.7228 
2.69^5 
2.6746 
2.65 1 1 
2.6279 


.2309 


4-3315 


•2339 
.2370 
.2401 
.2432 
.2462 


4.2747 
4-2193 
4-1653 
4.1126 
4.061 1 


.10510 


9-5144 


•3839 


2.6051 


.10805 
.11099 

•II394 
.11688 
.11983 


9-2553 
9.0098 
8.7769 
8-5555 
8.3450 


•3872 
•3906 
•3939 
-3973 
.4006 


2.5826 

2.5172 
2.4960 


•2493 


4.0108 


.2524 

■.IWe 

.2617 
.2648 


3.9617 
3-9136 
3.8667 
3.8208 
3.7760 


.12278 


8.1443 


.4040 


2-4751 


•12574 
.12869 
.13165 


7^9530 
7.7704 
7-5958 


-4074 
.4108 
.4142 


2.4545 
2.4342 
2.4142 


.2679 


3-7321 




Cot. 


Tan. 


A. 


Cot. 


Tan. 


A. 




Cot. 1 Tan. 


A. 



NATURAL TANGENTS AND COTANGENTS. 



23 



A. 


Tan. 


Cot. 




A. 


Tan. 


Cot. 




A. 


Tan. 


Cot. 




40' 
50' 
23° 


4142 
4176 
4210 


2.4142 

2.3945 
2.3750 


30; 

20' 
10' 

67° 
50; 
40' 
30' 
20' 
10' 

50; 
40' 
30; 
20' 
10' 
65° 
50; 
40' 
30; 
20' 
10' 
64° 

50; 
40' 
30; 
20' 
lo' 
63° 

50^ 
40 
30 
20' 
10' 
62° 
50; 
40' 
30' 
20' 
10' 
61° 

50' 
40' 
30' 
20' 
10' 
60° 


30° 

10' 

20' 
30; 
40' 
50' 
31° 
10' 
20' 
30' 
40' 
50' 
32° 
10' 
20' 

30; 
40' 

50' 
33° 
10' 
20' 
30' 
40' 
50' 
34° 
10' 
20' 
30' 
40' 
50/ 

35° 

10' 
20' 
30; 
40' 
50' 
36° 
10' 
20' 
30; 
40' 
50' 
37° 
10' 
20' 
30^ 


.5774 


I.7321 


60° 

50; 
40' 
30; 
20' 
10' 
59° 
50; 
40' 
30; 
20' 
10' 

58° 

50; 
40' 
30' 
20' 
10' 
57° 
50; 
40' 

30; 
20' 
10' 
56° 

50; 
40' 
30' 
20' 
10' 
55° 
50; 
40 
30 
20' 
10' 
54° 
50; 
40' 
30' 
20' 
10' 
53° 
50; 
40' 
30' 


30' 
40' 
50' 
38° 
10' 
20' 
30; 
40 
50' 
39° 
10' 
20' 
30' 
40' 

50' 
40° 
10' 
20' 
30; 
40' 
50' 
41° 
10' 
20' 
30; 
40' 
50' 
42° 
10' 
20' 
30' 
40' 
50' 
43° 
10' 
20' 
30; 
40 
50' 
44° 
10' 
20' 
30; 
40' 
50' 
45° 


-7673 
.7720 
.7766 


1.3032 
1.2954 
1.2876 


30; 
20' 
10' 

52° 
50; 
40' 
30; 
20' 
10' 
51° 
50; 
40' 
30' 
20' 
10' 
50° 
50; 
40' 
30; 
20' 
10' 
49° 
50; 
40' 
30' 
20' 
10' 

48° 

50; 
40/ 
30' 
20' 
10' 
47° 
50; 
40' 

30; 
20' 
10' 
46° 

50' 
40' 
30' 
20' 
10' 
45° 


.5812 
.5851 
.5890 
•5930 
.5969 


1.7205 
1.7090 
1.6977 
1.6864 
1-6753 


4245 


2.3559 


•7813 


1.2799 


10' 

20' 
30' 
40; 
50' 


4279 
4314 
4348 
4383 
4417 


2.3369 
2.3183 
2.2998 
2.2817 
2.2637 


.7860 
.7907 

•7954 
.8002 
.8050 


1.2723 
1.2647 
1.2572 
1.2497 
1.2423 


.6009 


1.6643 


.6048 
.6088 
.6128 
.6168 
.6208 


1 .6426 

I.63I9 
I.62I2 
1. 6107 


24° 


4452 


2.2460 


.8098 


1.2349 


10' 
20' 
30; 
40/ 

50' 


4487 
4522 

4557 
4592 
4628 


2.2286 
2.2113 

2.1943 
2.1775 
2.1609 


.8146 
.8195 
-8243 
.8292 

.8342 


1.2276 
1.2203 
1-2131 
1.2059 
1. 1988 


.6249 


1.6003 


.6289 

.6330 

•6371 
.6412 

•6453 


1.5900 
1-5798 
1.5697 

1-5597 
1-5497 


25° 


4663 


2.1445 


.8391 


I.1918 


10' 
20' 
30; 
40' 
50' 


4699 
4734 

4806 
4841 


2.1283 
2.II23 

2.0965 
2.0809 
2.0655 


.8441 
.8491 
.8541 
.8591 
.8642 


1. 1847 
1. 1778 
1 . 1 708 
1. 1640 
I-I571 


.6494 


1-5399 


.6536 

nil 
.6661 
.6703 


I -5301 
1.5204 
1. 5108 
1-5013 
1-4919 


26° 


4877 


2.0503 


.8693 


1. 1504 


10' 
20' 
30' 
40; 
50' 


4913 
4950 
4986 
5022 
5059 


2.0353 
2.0204 
2.0057 
I.9912 

1.9768 


.8744 
.8796 
.8847 
.8899 
.8952 


1.1436 
1. 1369 

1. 1303 
1. 1237 
I.II71 


•6745 


1.4826 


.6787 
.6830 

-6873 
.6916 

•6959 


1-4733 
1. 464 1 

1-4550 
1 .4460 

1-4370 


27° 


5095 


1.9626 


.9004 


I.II06 


10' 
20' 
30; 
40' 
50' 


5132 
5169 
5206 

5243 
5280 


1.9486 

1-9347 
1.9210 
1.9074 
1 .8940 


•9057 
.9110 
.9163 
.9217 
.9271 


I.IO41 
1.0977 

1-0913 
1.0850 
1.0786 


.7002 


r.4281 


.7046 
.7089 

-7133 

.7177 
.7221 


1-4193 
1.4106 
1. 4019 

1-3934 

1.3848 


28° 


5317 


1.8807 


-9325 


1.0724 


10' 
20' 
30' 
40' 
50' 


5354 
5392 
5430 
5467 
5505 


1.8676 
1.8546 
1.8418 
1.8291 
1.8165 


.9380 

-9435 
-9490 

•9545 
.9601 


1. 066 1 
1.0599 
1.0538 
1.0477 
I.0416 


.7265 


1-3764 


•7310 

•7355 
.7400 

-7445 
.7490 


1.3680 
1-3597 
1-3514 
1-3432 
1-3351 


29° 


5543 


1.8040 


.9657 


1-0355 


10' 
20' 
30' 
40; 
50' 


5581 
5619 
5658 
5696 

5735 


1.7917 
1.7796 

1-7675 
1-7556 
1-7437 


•9713 
.9770 
.9827 
.9884 
.9942 


1.0295 
1.0235 
I.0176 
I.0117 
1.0058 


•7536 


1.3270 


.7581 
.7627 
•7673 


1.3190 
1.3111 
1.3032 


30° 


5774 


1.7321 


I.OOOO 


1 .0000 




Cot. 


Tan. 


A. 




Cot. 


Tan. 


A. 




Cot. 


Tan. 


A. 



<: 



LBFe'26 



